NCERT Solutions: Area

Page No. 148 Math Talk

Q: Try to think of different creative ways to divide a square into 4 parts of equal area.

Ans: There are infinitely many ways to divide a square into 4 parts of equal area. Here are some examples:

• Method 1: Draw two perpendicular lines through the center of the square, dividing it into 4 equal smaller squares.
• Method 2: Draw two diagonal lines connecting opposite corners, creating 4 equal triangles.
• Method 3: Draw parallel lines dividing the square into 4 equal horizontal or vertical strips.
• Method 4: Use curved lines – draw arcs from corners that meet at the center, creating 4 equal curved regions.

Page No. 150

Q : Why do we count the number of unit squares to assign measures for area? Couldn’t we have just used the perimeter of a region as a measure of its area?

Ans: We count unit squares to measure area because area depends on the surface covered, not just the boundary.
Perimeter only measures the length of the boundary, not how much region is enclosed.
Two regions can have the same perimeter but different areas, so perimeter cannot correctly represent area.

Q: If two regions have the same perimeter, can’t we conclude that they have the same area? Or, if one region has a larger perimeter than another, can’t we conclude that it also has a larger area?

Ans: No, we cannot conclude that.

• Two regions may have the same perimeter but different areas.
• A region can have a larger perimeter but smaller area than another region.

Hence, perimeter is not indicative of area.

Q: Find two rectangles that are examples of such regions. If needed, use a grid paper (given at the end of the book) for this.

Ans: Do it Yourself!

Math Talk

Q: Also, give an example of two regions of other shapes where the region with the larger perimeter has the smaller area! This property should be visually clear in your example.

Ans: Consider:

• A long, thin zig-zag shaped region
• A compact circular region

The zig-zag shape has a very large boundary (perimeter) but covers very little area.
The circle has a smaller perimeter but covers a much larger area.

Thus, a region with a larger perimeter can have a smaller area, which is visually clear.

Figure it Out

Q1: Identify the missing side lengths.

Ans: (i) After naming the figureIn rectangle ABCD,
Area of rectangle = Length × Breadth
7 × BC = 21
⇒ BC = 3 in

∴ AF = AD + DF
= 3 in + 4 in
= 7 in
In the rectangle EFAG,
EF × AF = 28 in²
⇒ EF × 7 in = 28 in²
⇒ EF = 4 in
∴ HA = HG + GA
= 3 in + 4 in
= 7 in
In rectangle HIJA,
Area = HA × AJ
⇒ 35 in² = 7 in × AJ
⇒ AJ = 5 in
∴ AK = AJ + JK
= 5 in + 2 in
= 7 in
In rectangle KLMA,
Area = KL × LM
⇒ x in × 7 in = 14 in²
⇒ x = 2 in

Thus, the missing sidelength = 2 in

(ii) After naming the figure,

In rectangle ABGH,
Area = AB × AH
AB × 4 m = 29 m²
AB = 29/4 m or 7.25 m
Area of rectangle HGDC = Area of rectangle ABDC – Area of rectangle ABGH
= 50 m² – 29 m²
= 21 m²
In rectangle HGDC,
CD × GD = 21 m²
⇒ 29/4m × GD = 21 m²
⇒ GD = 84/29 m or 2.9 m
In rectangle BEFG,
BG × BE = 11 m²
⇒ 4 m × M = 11 m²
⇒ BE = 11/4m or 2.75 m
Thus, AB = 29/4 m; BE = 11/4 m and GD = 84/29 m

Page No. 151

Q2: The figure shows a path (the shaded portion) laid around a rectangular park EFGH.

(i) What measurements do you need to find the area of the path? Once you identify the lengths to be measured, assign possible values of your choice to these measurements and find the area of the path. Give a formula for the area. [Hint: There is a relation between the areas of EFGH, the path, and ABCD.]

Ans: Measurements needed:
Length of outer rectangle ABCD = A
Width of outer rectangle ABCD = B
Length of inner rectangle EFGH = a
Width of inner rectangle EFGH = b
Let A = 10, B = 8, a = 6, b = 4.
Calculation: Area of path = Area of ABCD – Area of EFGH
= 10 × 8 – 6 × 4
= 80 – 24
= 56 m²
Formula:
Area of path = (A × B) – (a × b)

(ii) If the width of the path along each side is given, can you find its area? If not, what other measurements do you need? Assign values of your choice to these measurements and find the area of the path. Give a formula for the area using these measurements. [Hint: Break the path into rectangles.]

Ans: Yes, if the width of the path is uniform along each side, we can find its area, but we also need the dimensions of either the outer or inner rectangle.
If the width of the path d = 2 m (Uniform on all sides)
Length of inner path EFGH = l = 16 m
Width of inner path EFGH = w = 11 m
Length of outer rectangle = l + 2d
= 16 + 2(2)
= 20 m
Width of outer rectangle = w + 2d
= 11 + 2(2)
= 15 m
Breaking the path into rectangles
Now there are 4 rectangles
Left rectangle = w × d
= 11 × 2
= 22 m²
Right rectangle = w × d
= 11 × 2
= 22 m²
Top rectangle = (l + 2d) d
= 20 × 2
= 40 m²
Bottom rectangle = (l + 2d) d
= 20 × 2
= 40 m²
Total area of path = 22 + 22 + 40 + 40 = 124 m²
Formula:
Area of path = 2d(l + w) + 4d
where d – width of path
l – length of inner path
w – width of inner path

(iii) Does the area of the path change when the outer rectangle is moved while keeping the inner rectangular park EFGH inside it, as shown?

Ans: No, the area of the path does not change.
Reason: The area of the path depends only on:
Area of outer rectangle ABCD.
Area of inner rectangle EFGH.

Math Talk

Q3: The figure shows a plot with sides 14 m and 12 m, and with a crosspath. What other measurements do you need to find the area of the crosspath? Once you identify the lengths to be measured, assign some possible values of your choice and find the area of the path. Give a formula for the area based on the measurements you choose.

Ans: Measurements needed:
Length of plot = 14 m
Width of plot = 12 m
Width of horizontal path = x₁
Width of vertical path = x₂
Assign values:
Let x₁ = 2 m
x₂ = 2 m
Now area of horizontal path = 14 × 2 = 28 m
Area of vertical path = 12 × 2 = 24 m
Area of overlapping square = 2 × 2 = 4 m²
∴ Area of cross path = 28 + 24 – 4 = 48 m²
Formula = Area of cross path = (L × w₁) + (W × w₂) – (w₁ × w₂)
Here, L = length of plot
W = Width of plot
w₁ = Width of horizontal path
w₂ = Width of vertical path

Page No. 152

Q4: Find the area of the spiral tube shown in the figure. The tube has the same width throughout.

[Hint: There are different ways of finding the area. Here is one method.]

What should be the length of the straight tube if it is to have the same area as the bent tube on the left?

Ans: After naming the figure,

The area of the spiral tube = Area of the rectangle, ABEC + Area of the rectangle, DEGF + Area of the rectangle, GHIJ + Area of the rectangle, JKML + Area of the rectangle, NOPL + Area of the rectangle, PQRS + Area of the rectangle, STUV + Area of the rectangle, VWYX + Area of the rectangle, XZA₁B₁
= AC × AB + EG × DE + IH × JI + LJ × LM + NO × NL + PQ × PS + UT × ST + VX × VW + ZA₁ × A₁B₁
= 20 × 1 + 18 × 1 + 20 × 1 + 13 × 1 + 15 × 1 + 8 × 1 + 10 × 1 + 3 × 1 + 5 × 1
= 20 + 18 + 20 + 13 + 15 + 8 + 10 + 3 + 5
= 112 sq. units
Thus, the area of the spiral tube is 112 sq. units.
Let the length of the straight tube be x.
The area of the bent tube on the left = Area of rectangle, BACD + Area of rectangle, BGFE
= AC × CD + BG × BE
= 5 × 1 + 4 × 1
= 9 sq. units

Area of straight tube = x × 1
Area of bent tube = 9 sq. units
According to the question, both are the same.
So x × 1 = 9
⇒ x = 9 units

Q5: In this figure, if the sidelength of the square is doubled, what is the increase in the areas of the regions 1, 2 and 3? Give reasons.

Ans: Let ‘a’ be the side of the square.

Area of triangle, DCB (region 3)

In right angled DCB,

Area of triangle, AED (region 1)

Area of triangle, AEB (region 2)

If the side of the square is doubled = 2a

Area of triangle, BCD (region 3)

In right angled triangle, BCD,
By Pythagoras theorem,

Area of triangle, AED (region 1)

Area of triangle, AEB (region 2)

The increase in area of region 1

The increase in area of region 2

The increase in area of region 3

Thus, the increase in the areas of regions 1, 2, and 3 is 4 times.

Reason: If the sidelength of the square is doubled, then the area becomes 4 times.

Math Talk

Q6: Divide a square into 4 parts by drawing two perpendicular lines inside the square as shown in the figure. Rearrange the pieces to get a larger square, with a hole inside.

Ans: 1. Let us take a square of cardboard (8 cm × 8 cm).
2. Draw two perpendicular lines inside the square (not through the center), dividing it into 4 rectangular pieces.
3. Cut along these lines to get 4 pieces.
4. Rearrange these 4 pieces.
Place them at the four corners of a larger imaginary square.
The pieces should be arranged so that they form a square with a hole in the middle.

Page No. 153

Q: In the given figure, which triangle has a greater area: ∆XDC or ∆YDC, if both the rectangles are identical?

Ans: Both triangles have equal area.

Reason:

• Both rectangles ABCD are identical
• ∆XDC has base DC and height from X to DC (which is the height of rectangle AB)
• ∆YDC has base DC and height from Y to DC (which is also the height of rectangle AB)

Since both triangles have the same base (DC) and the same height (height of rectangle), they have equal areas.

Area of ∆XDC = ½ × DC × h = Area of ∆YDC

where h is the height of the rectangle.

Each triangle has exactly half the area of rectangle ABCD.

Q: In the given figure, which triangle has a greater area: ∆XDC or ∆YBC, if both the rectangles are identical?

Ans: Both triangles have equal area.

Reason:

• Both rectangles are identical
• ∆XDC in the first rectangle and ∆YBC in the second rectangle
• By dropping altitudes from X and Y, we can see that both triangles occupy exactly half the area of their respective rectangles
• Since the rectangles are identical, half of each rectangle’s area is the same

Therefore, Area of ∆XDC = Area of ∆YBC = ½ × (Area of rectangle ABCD)

Q: Find the area of ∆XDC.

Ans: Given from the figure:

• Rectangle has dimensions 5 units × 4 units
• Point X lies on side AB
• ∆XDC is formed

Method 1: Using rectangle area Area of rectangle ABCD = $5 \times 4$ = 20 square units

Since diagonal or any line from a vertex on one side to a point on the opposite side divides areas: Area of ∆XDC = ½ × base × height = ½ × DC × AB = ½ × $5 \times 4$ = 10 square units

Method 2: Direct formula Taking DC as base = 5 units Height from X perpendicular to DC = 4 units (height of rectangle)

Area = ½ × $5 \times 4$ = 10 square units

Page No. 157Figure it Out

Q1: Find the areas of the following triangles:

Ans: (i) Area of triangle ABC = 1/2 × base × height

= 1/2 × BC × AE

= 1/2 × 4 cm × 3 cm

= 6 cm²

Thus, the area of the triangle, ABC = 6 cm²

(ii) Area of triangle DEF = 1/2 × EF × ND

= 1/2 × 5 cm × 3.2 cm

= 5 × 1.6 cm²

= 8 cm²

Thus, the area of the triangle DEF = 8 cm²

(iii) Area of triangle = 1/2 × base × height

Area of ΔNAT = 1/2 × AT × NA

= 1/2 × 3 cm × 4 cm

= 6 cm²

Thus, the area of the triangle, NAT = 6 cm²

 Q2: Find the length of the altitude BY.

Ans: Area of ΔAXC = 1/2 × XC × AX

= 1/2 × (XB + 6) × 4

= 2 × (XB + 6)

= 2 × XB + 12 sq. units

Area of ΔAXB = 1/2 × XB × AX

= 1/2 × XB × 4

= 2 × XB sq. units

Area of ΔABC = 1/2 × AC × BY

= 1/2 × 8 × BY

= 4BY

∴ Area of ΔAXC = Area of ΔAXB + Area of ΔABC

2XB + 12 = 2XB + 4BY

⇒ 4BY = 12

⇒ BY = 3 units

Thus, the length of the altitude BY is 3 units.

Q3: Find the area of ∆SUB, given that it is isosceles, SE is perpendicular to UB, and the area of ∆SEB is 24 sq. units.

Ans: Given,
The area of ∆SEB = 24 sq. units
Given that ∆SUB is an isosceles triangle.
SU and SB are equal sides, and UB is the base.
∴ SE is perpendicular to UB
⇒ UE = EB
SE is the common base of ∆SUE and ∆SEB.
∴ Area of ∆SEB = 24 sq. units = Area of ∆SEU
∴ The area of triangle SUB = Area of ∆SEU + Area of ∆SEB
= 24 + 24
= 48 sq. units
Thus, the area of ∆SUB is 48 sq. units.

Q4: [Śulba-Sūtras] Give a method to transform a rectangle into a triangle of equal area.

Ans: 1. Let us take a rectangle ABCD, with length a and breadth b.

2. Now mark the midpoint E of side CD.
Draw a line perpendicular to CD passing through E.
Mark a point M on it such that ME = b.
3. Draw a triangle using base = AB (same as the rectangle’s length) and join M to A and B.

Q5: [Śulba-Sūtras] Give a method to transform a triangle into a rectangle of equal area.

Ans: 1. Take a triangle ABC with base b and height h.

2. Find the midpoint M of the height.
3. Draw a line parallel to the base through M.
This line intersects the sides of the triangle.
4. Create a rectangle using.
Length = Same as the triangle’s base = b
Width = Half of the triangle’s height = h/2h2

Q6: ABCD, BCEF, and BFGH are identical squares.

(i) If the area of the red region is 49 sq. units, then what is the area of the blue region?

Ans: Given that ABCD, BCEF, and BFGH are identical.
The area of the red region (∆HBI + IBCD) = 49 sq. units.
Let the side of each square be a.
∴ IB = AB/2 = a/2 units
Let ‘a’ units be the side of the square.
(i) Area of the red region ∆HDC = 1/2 × DC × DC
(∴ HC = HB + BC = a + a = 2a)
= 1/2 × a × 2a
⇒ 1/2 × 2a² = 49
⇒ a = 7 units

∴ The area of the black region, ∆IAD = 1/2 × AI × AD
= 1/2 × 7/2 × 7
= 49/4 sq. units
= 12.25 sq. units
Thus, the area of the black region is 12.25 square units.

(ii) In another version of this figure, if the total area enclosed by the blue and red regions is 180 sq. units, then what is the area of each square?

Ans: Given, the total area enclosed by the black and red regions = Area of ΔHDC + Area of ΔAID = 180 sq. units

Let ‘a’ be the side of the square.

∴ The area of each square = a²
= (12)²
= 144 sq. units
Thus, the area of each square is 144 sq. units

Page No. 159Try This

Q7: If M and N are the midpoints of XY and XZ, what fraction of the area of ∆XYZ is the area of ∆XMN? [Hint: Join NY]

Ans: Let O be the midpoint of YZ, then join M to O and N to O.

According to mid point theorem,
MN = 1/2 YZ, and MN is parallel to YZ.
The triangle XYZ is divided into four equal triangles.
So, Area of ∆XMN = 1/4 × Area of ∆XYZ.

Math Talk

Q8: Gopal needs to carry water from the river to his water tank. He starts from his house. What is the shortest path he can take from his house to the river and then to the water tank? Roughly recreate the map in your notebook and trace the shortest path.

Ans: Let P be the point on the bank of the river, then the shortest path from H to P to T is created by the point P such that ∆HPT has minimum area.

Shortest path
House (H) → P (Point on river) → Water tank

Q: How do we find the area of this pentagon?

Ans: To find the area of the given pentagon, we divide it into triangles by drawing diagonals from one vertex to the other non-adjacent vertices.

Each triangle’s area is found using the formula:

$=\frac{1}{2}\times \times$Area of a triangle=21×base×height

After finding the areas of all the triangles, we add them together to get the area of the pentagon.

Thus, the area of a pentagon (or any polygon) can be found by:

• Dividing it into triangles
• Finding the area of each triangle
• Adding all the triangle areas

Page No. 160Figure it Out

Q1: Find the area of the quadrilateral ABCD given that AC = 22 cm, BM = 3 cm, DN = 3 cm, BM is perpendicular to AC, and DN is perpendicular to AC.

Ans: Area of the quadrilateral ABCD = Area of triangle CAD + Area of triangle ACB

Area of ∆ACB = 1/2 × AC × BM

= 1/2 × 22 cm × 3 cm

= 33 cm²

Area of ∆CAD = 1/2 × AC × DN

= 1/2 × 22 cm × 3 cm

= 33 cm²

∴ The area of the quadrilateral ABCD = 33 cm² + 33 cm² = 66 cm²

Q2: Find the area of the shaded region given that ABCD is a rectangle.

Ans: The area of the shaded region = Area of the rectangle ABCD – (Area of triangle AEF + Area of triangle EBC)

Area of the rectangle ABCD = Length × Breadth

= AB × AD

= 18 cm × 10 cm [∵ AB = AE + EB = 10 cm + 8 cm = 18 cm; AD = AF + FD = 6 cm + 4 cm = 10 cm]

= 180 cm²

Area of the triangle AEF = 1/2 × AE × AF

= 1/2 × 10 cm × 6 cm

= 30 cm²

Area of the triangle EBC = 1/2 × EB × BC

= 1/2 × 8 cm × 10 cm

= 40 cm²

∴ The area of the shaded region = 180 cm² – (30 cm² + 40 cm²)

= 180 cm² – 70 cm²

= 110 cm²Math Talk

Q3: What measurements would you need to find the area of a regular hexagon?

Ans: Minimum measurement needed

Side length (l) of the hexagon

Area of regular hexagon = (3√3 a²)/2

where l is the side length.Math Talk

Q4: What fraction of the total area of the rectangle is the area of the blue region?

Ans: Let l be the length and b be the breadth of the rectangle ABCD.

Total area of rectangle, ABCD = DC × BC = l × b sq. units

Area of ΔAOB = 1/2 × AB × OE = 1/2 × l × x sq. units

Area of ΔDOC = 1/2 × DC × OF = 1/2 × l × y sq. units

∴ The area of the red region = Area of ΔAOB + Area of ΔDOC

= 1/2 × l × x + 1/2 × l × y

= 1/2 × l × (x + y) sq. units

= 1/2 × l × b sq. units  [∵ x + y = b]

∴ Area of red region = 1/2 × Area of rectangle

Thus, the required fraction is 1/2Math Talk

Q5: Give a method to obtain a quadrilateral whose area is half that of a given quadrilateral.

Ans: Let ABCD be a given quadrilateral.
Mark mid points of AB, BC, CD, and DA as P, Q, R, and S.
Join midpoints, then PQRS is the required quadrilateral with half the area of the given quadrilateral ABCD.Page No. 162-163Figure it Out

Q1: Observe the parallelograms in the figure below.

(i) What can we say about the areas of all these parallelograms?

Ans: (i) (a) Area of parallelogram = base × height
= 5 × 3
= 15 sq. units
(b) Area of parallelogram = 5 × 3 = 15 sq. units
(c) Area of parallelogram = 5 × 3 = 15 sq. units
(d) Area of parallelogram = 5 × 3 = 15 sq. units
(e) Area of parallelogram = 5 × 3 = 15 sq. units
(f) Area of parallelogram = 5 × 3 = 15 sq. units
(g) Area of parallelogram = 5 × 3 = 15 sq. units
All parallelograms have equal areas.

(ii) What can we say about their perimeters? Which figure appears to have the maximum perimeter, and which has the minimum perimeter?

Ans: (ii) The perimeters of these parallelograms are different even though their areas are the same.
Figure (d) has the minimum perimeter, and Figure (g) has the maximum perimeter.Page No. 163

Q2: Find the areas of the following parallelograms:

Ans: Area of the parallelogram = base × height
(i) Here, base = 7 cm and height = 4 cm
Area of the parallelogram = 7 cm × 4 cm = 28 cm²
(ii) Here, base = 5 cm and height = 3 cm
Area of the parallelogram = 5 cm × 3 cm = 15 cm²
(iii) Here, base = 5 cm and height = 4.8 cm
Area of the parallelogram = 5 cm × 4.8 cm = 24 cm²
(iv) Here, base = 2 cm and height = 4.4 cm
Area of the parallelogram = 2 cm × 4.4 cm = 8.8 cm²

Q3: Find QN.

Ans: In ∆PNQ, ∠PNQ = 90°
PN = 7.6 cm
PQ = 12 cm
By Pythagoras theorem
PQ² = PN² + NQ²
⇒ (12)² = (7.6)² + NQ²
⇒ QN² = (12)² – (7.6)²
⇒ QN² = 144 – 57.76
⇒ QN² = 86.24
⇒ QN = 9.28 cm

Q4: Consider a rectangle and a parallelogram of the same sidelengths: 5 cm and 4 cm. Which has the greater area? [Hint: Imagine constructing them on the same base.]

Ans: For rectangle:
l = 5, w = 4, all angles = 90°
∴ Area = l × w
= 5 × 4
= 20 cm²
For parallelogram:
Base (b) = 5 cm, one slanted side = 4 cm
Height will be less than 4 cm because the side is slanted.
Area = 5 × 4 < 20 cm²
Hence, the rectangle has a greater area than the parallelogram.

Q5: Give a method to obtain a rectangle whose area is twice that of a given triangle. What are the different methods that you can think of?

Ans: Given: Triangle with area A
Required: Rectangle with area = A
Method 1: If the triangle has base b and height h
Area of triangle = 1/2 × b × h = A
To get a rectangle with an area of 2A.
Take length = b, width = h
Area of a rectangle = b × h
= 2 × (1/2 × b × h)
= 2A
Steps:
1. Measure the base and height of the given triangle.
2. Construct a rectangle with these measurements as length and width.
Method 2: Scaling method:
1. Take the rectangle.
2. Create a rectangle with base = (base of triangle) and height = height of triangle.
3. This rectangle automatically has twice the area of the rectangle.Page No. 164

Q6: [Śulba-Sūtras] Give a method to obtain a rectangle of the same area as a given triangle.

Ans: Given: A triangle with base b and height h.
Required: Rectangle with the same area
Area of triangle = 1/2 b h
To get a rectangle with the same area
Rectangle length = b/2(half the triangle’s base)
Rectangle width = h (same as the triangle’s height)
Area = b/2 × h = 1/2 × b × h

Q7: [Śulba-Sūtras] An isosceles triangle can be converted into a rectangle by dissection in a simpler way. Can you find out how to do it?

[Hint: Show that triangles ADB and AADC can be made into halves of a rectangle. Figure out how they should be assembled to get a rectangle. Use cut-outs if necessary.]

Ans: Given: Isosceles triangle ABC, where AB = AC, and AD is the altitude from A to BC.
Method: Since the triangle is isosceles:
AD is perpendicular to BC. D is the midpoint of BC (property of an isosceles triangle).
AD bisects the triangle into two congruent right triangles: ∆ADB and ∆ADC.
Dissection Process:
Step 1: The altitude AD divides the isosceles triangle into two congruent right triangles ∆ADB and ∆ADC.
Step 2: Each of these right triangles can be made into half of a rectangle.
Step 3: Assembly:
Take triangle ∆ADB
Take triangle ∆ADC
Rotate one triangle 180°
Arrange them so that:
The two equal sides (AB and AC) form opposite sides of a rectangle.
The altitude AD appears twice, forming the other pair of opposite sides.
Step 4: The resulting figure is a rectangle with:
Length = BC (base of the isosceles triangle)
Width = AD/2 (half the altitude) or alternatively.
Length = AB (= AC, the equal sides)
Width related to the base.

Q8: [Śulba-Sūtras] Give a method to convert a rectangle into an isosceles triangle by dissection.

Ans: This is the reverse of Question 7.

Method:

Given: Rectangle PQRS with length l and width w.

​Required: Isosceles triangle with the same area
Dissection Steps:
Step 1: Take a rectangle PQRS with PQ = l and PS = w.
Step 2: Mark the midpoint M of side PQ.
Step 3: From M, draw lines to the bottom corners R and S.
Step 4: Cut the rectangle into three pieces:
Triangle PMS (left); Triangle QMR (right); Central region (if any)
Step 5: Rearrange:
Flip the triangle PMS and attach it along MS to form one half of the triangle.
Flip triangle QMR and attach it along MR to form the other half.
These create an isosceles triangle.

Q9: Which has greater area — an equilateral triangle or a square of the same sidelength as the triangle? Which has greater area — two identical equilateral triangles together or a square of the same sidelength as the triangle? Give reasons.

Ans: Area of equilateral triangle = (√3 / 4) a²

Area of square = a²

⇒ (√3 / 4) a² < a²

So, the area of a square is greater than the area of an equilateral triangle of the same side length.

Area of two identical equilateral triangles = (√3 / 4) a² + (√3 / 4) a²

= (2√3 / 4) a²

= (√3 / 2) a²

Area of square of side length a = a²

Clearly, (√3 / 2) a² < a²

So, the area of a square is greater than the area of two identical equilateral triangles.Page No. 169Figure it Out

Q1: Find the area of a rhombus whose diagonals are 20 cm and 15 cm.

Ans: Given, first diagonal = 20 cm

second diagonal = 15 cm

The area of a rhombus = 1/2 × (Product of diagonals)

= 1/2 × First diagonal × second diagonal

= 1/2 × 20 cm × 15 cm

= 150 cm²

Thus, the area of a rhombus is 150 cm²

Q2: Give a method to convert a rectangle into a rhombus of equal area using dissection.

Ans: This is the reverse of the rhombus to rectangle dissection.
Method:
Given: Rectangle PQRS with length l and width W
Required: Rhombus with the same area = l × W
Dissection Process:
Step 1: The rhombus will have diagonals d₁ and d₂ such that: 1/2 × d₁ × d₂ = l × w
So, d₁ × d₂ = 2lw.
Step 2: Choose convenient diagonal lengths:
Let d₁ = 2l (twice the rectangle length)
Then d₂ = w (same as rectangle width)
Check: 1/2 × 2l × w = lw
Or
Let d₁ = 2w (twice the rectangle width)
Then d₂ = l (same as rectangle length)
Step 3: Dissection process (reverse of textbook method):
Divide the rectangle into two halves
Mark the center point O.
Cut and rotate pieces to form two isosceles triangles.
Arrange these triangles to share a common diagonal.
This creates a rhombus. 

Q3: Find the areas of the following figures:

Ans: The area of the trapezium = 1/2 × (Sum of parallel sides) × (Distance between them) = 1/2 × (a + b) × h

(i) Here, a = 10 ft, b = 7 ft and h = 16 ft

Area of trapezium = 1/2 × (10 + 7) × 16

= 17 × 8

= 136 ft²

(ii) Here, a = 36 m, b = 24 m and h = 14 m

Area of trapezium = 1/2 × (36 + 24) × 14

= 60 × 7

= 420 m²

(iii) Here, a = 14 in, b = 6 in and h = 10 in

Area of trapezium = 1/2 × (14 + 6) × 10

= 20 × 5

= 100 in²

(iv) Here, a = 18 ft, b = 12 ft and h = 8 ft

Area of trapezium = 1/2 × (18 + 12) × 8

= 30 × 4

= 120 ft²

Q4: [Śulba-Sūtras] Give a method to convert an isosceles trapezium to a rectangle using dissection.

Ans: An isosceles trapezium has special properties that make dissection simpler.

​Properties of Isosceles Trapezium ABCD:
AB || CD (parallel sides)
AD = BC (non-parallel sides are equal)
∠A = ∠B and ∠D = ∠C (base angles are equal)
Dissection Method:
Step 1: Draw perpendiculars from C and D to AB, meeting at points P and Q, respectively.
This creates rectangle PQDC in the middle.
Two congruent right triangles: ΔAPD and ΔBQC.
Step 2: Since the trapezium is isosceles:
ΔAPD ≅ ΔBQC (congruent triangles)
AP = BQ
Step 3: Rearrangement:
Cut triangle ΔAPD
Rotate it and attach it to the right side (next to ΔBQC)
The two triangles together form a rectangle with a width = height of the trapezium
Step 4: Combine: The central rectangle PQDC
The rectangle formed from the two triangles.
These can be joined to form one large rectangle.
Resulting Rectangle:
Length = CD + AP
= CD + (AB – CD)/2
= (AB + CD)/2
Width = h (height of trapezium)
Area = (AB + CD)/2 × h = 12h(AB + CD)
This matches the trapezium area formula!

Q5: Here is one of the ways to convert trapezium ABCD into a rectangle EFGH of equal area. Given the trapezium ABCD, how do we find the vertices of the rectangle EFGH?

[Hint: If AАНI = ADGI and ABEJ =ACFJ, then the trapezium and rectangle have equal areas.]

Ans: Given: Trapezium ABCD with AB || CD
Required: Find the positions of the vertices E, F, G, and H to form a rectangle EFGH with equal area
Using the Hint: If ΔAHI ≅ ΔDGI and ΔBEJ ≅ ΔCFJ, then areas are equal.
Method:
Step 1: The rectangle EFGH should have:
EF as one side (top side)
GH is the opposite parallel side (bottom side)
EH and FG are the other pair of sides.
Step 2: Position the rectangle such that:
Points I and J are strategically chosen on the trapezium.
ΔAHI (part outside rectangle on left) is cut and moved to fill ΔDGI.
ΔBEJ (part outside rectangle on right) is cut and moved to fill ΔCFJ.
Step 3: For congruency:
Mark I on side AD
Mark J on side BC
Choose positions such that:
HI = GI (making ΔAHI ≅ ΔDGI possible)
EJ = FJ (making ΔBEJ ≅ ΔCFJ possible)
Step 4: The height of the rectangle = h (height of the trapezium)
Step 5: The length of rectangle = (a + b)/2
where a and b are parallel sides
This ensures: Area of rectangle = h × (a + b)/2 = Area of trapezium
Practical construction:
Draw the trapezium ABCD
Calculate required rectangle length = (AB + CD)/2
Mark points H and E on AB such that the central portion has this length.
Draw perpendiculars to get rectangle EFGH.
Verify that the triangular pieces outside match those inside.
This construction beautifully demonstrates area conservation through dissection!Page No. 170Math Talk

Q6: Using the idea of converting a trapezium into a rectangle of equal area, and vice versa, construct a trapezium of area 144 cm².

Ans: Area of the trapezium = 1/2 × (10 + 8) × 16 = 144 cm²
Area of square = 16 × 9 = 144 cm²

Q7: A regular hexagon is divided into a trapezium, an equilateral triangle, and a rhombus, as shown. Find the ratio of their areas.

Ans: Here total area of hexagon = 6 × (√3 / 4) a² = (3√3 / 2) a²

Equilateral triangle:

Area = (√3 / 4) a²

Rhombus:

Area = 2 × (√3 / 4) a² = (√3 / 2) a²

Trapezium:

Remaining Area = Total area − Triangle area − Rhombus area

= (3√3 / 4) a² − (√3 / 4) a² − (√3 / 2) a²

= (3√3 / 4) a²

Ratio of areas = Triangle : Rhombus : Trapezium

= (√3 / 4) a² : (√3 / 2) a² : (3√3 / 4) a²

= 1 : 2 : 3

Q8: ZYXW is a trapezium with ZY ∥ WX. A is the midpoint of XY. Show that the area of the trapezium ZYXW is equal to the area of ∆ZWB.

Ans: ∠ZAY = ∠BAX (Vertically opposite angles)

​AY = AX (∵ A is mid point of XY)
∠YZB = ∠XBZ (∵ ZY || XB alternate interior angles are equal)
∠ZYA = ∠BXA (∵ they are alternate interior angles)
So ∆ZAY ≅ ∆BAX
By the AAA congruence.
So Area of ∆ZAY = Area of ∆BAY
Thus, Area of trapezium ZYXW = Area of triangle ZWBPage No. 170-171Areas in Real Life

Q: What do you think is the area of an A4 sheet? Its sidelengths are 21 cm and 29.7 cm. Now find its area.

Ans: We know that the size of an A₄ sheet is rectangular.
∴ The area of an A₄ sheet = Length × Breadth
= 21 cm × 29.7 cm
= 623.7 cm²

Q: Express the following lengths in centimeters:

(i) 5 in
​(ii) 7.4 in

We know that,
1 in = 2.54 cm
(i) 5 in = 5 × 2.54 cm = 12.7 cm
(ii) 7.4 in = 7.4 × 2.54 cm = 18.796 cm

Q: Express the following lengths in inches:

(i) 5.08 cm

(ii) 11.43 cm

Ans: We know that,
2.54 cm = 1 in
∴ 1 cm = 1/2.54 in
(i) 5.08 cm = 5.08 × 1/2.54 in = 2 in
(ii) 11.43 cm = 11.43 × 1/2.54 in = 4.5 in

Q: How many in² is 1 ft²?

Ans: We know that,
1 ft = 12 in
∴ 1 ft² = (12 in)²
= 12² in²
= 144 in²

Q: How many m² is a km²?

Ans: We know that,
1 km = 1000 m
1 km² = 1000 m × 1000 m
= 1000000 m²
= 10⁶ m²
Thus, 1 km² = 1000000 m²

Q: How many times is your village/town/city bigger than your school?

Ans: Do it yourself

Q: Find the city with the largest area in (i) India, and (ii) the world.

Ans: Do it Yourself.

Q: Find the city with the smallest area in (i) India, and (ii) the world.

Ans: (i) India:

• Smallest city by area: Mahe (Puducherry Union Territory)
• Area: approximately 9 km²

(ii) World:

• Smallest city/country: Vatican City
• Area: approximately 0.44 km² (44 hectares)

NCERT Solutions: Algebra Play

Page No. 136

Q: How would you change this game to make the final answer 3? What about 5?

Ans: (a) We can understand such tricks through algebra.
Step 1: Think of a number = x
Step 2: Triple it = 3x
Step 3: Add 9 = 3x + 9
Step 4: Divide by 3 = 3x+9/3 = x + 3
Step 5: Subtract the original number you thought of (x + 3) – x = 3.
For Example:
Consider a number 23.
Triple it = 3 × 23 = 69
Add 9 = 69 + 9 = 78
Divide by 3 = 78 ÷ 3 = 26
∴ 26 – 23 = 3

(b) To make the final answer 5.
Step 1: Think of a number = x
Step 2: Double it = 2x
Step 3: Add 10; 2x + 10
Step 4: Divide by 2 = 2x+10/2 = x + 5
Step 5: Subtract the original number = x + 5 – x = 5

Q: Can you come up with more complicated steps that always lead to the same final value?

Ans: Yes. Here is an example.
We can understand such tricks through algebra.
Step 1: Think of a number = x
Step 2. 5 times it = 5x
Step 3: Add 25 = 5x + 25
Step 4: Divide by 5 = 5x+25/5 = x + 5
Step 5: Subtract the original number you thought of (x + 5) – x = 5
For Example:
Consider a number 18.
5 times 18 = 90
Add 25 = 90 + 25 = 115
Divide by 5 = 115 ÷ 5 = 23
∴ 23 – 18 = 5

Page No. 137

Q: Find the dates if the final answers are the following:

(i) 1269

Ans: 1269 = 100 M + 165 + D
Here, M = Month, D = Day
1269 – 165 = 100 M + D
⇒ 1104 = 100 M + D
⇒ 1100 + 04 = 100 M + D
∴ M = 11, D = 04
Thus, the date is 4th of November, i.e., 04/11

(ii) 394

Ans: 394 = 100 M + 165 + D
⇒ 394 – 165 = 100 M + D
⇒ 229 = 100 M + D
⇒ 200 + 29 = 100 M + D
∴ M = 02, D = 29
Thus, the date is 29th of February, i.e., 29/02.

(iii) 296

Ans: (iii) 296 = 100 M + 165 + D
⇒ 296- 165 = 100 M + D
⇒ 131 = 100 M + D
⇒ 100 + 31 = 100 M + D
∴ M = 01, D = 31
Thus, the date is 31st of January, i.e., 31/01.

Page No. 138

Q: Use the same rule to fill these pyramids:

Ans:

(i) 8 = 6 + 2(ii) 3 + 4 = 7; 4 + 3 = 7 and 7 + 7 = 14

(iii) 5 + 4 = 9; 4 + 5 = 9 and 5 + 0 = 5;
9 + 9 = 18; 9 + 5 = 14 and 18 + 14 = 32.

Page No. 139

Q: Fill the following pyramids:

Ans: (i) Let a, b, c, d, e, and f be the missing numbers.

a + 22 = 50
⇒ a = 50 – 22
⇒ a = 28
b + c = 28 …..(i)
c + d = 22 …..(ii)
Adding equations (i) and (ii), we get
b + d + 2c = 50 ……(iii)
Also, 4 + e = b …..(iv)
e + 6 = c …(v)
6 + f = d ….(vi)
Adding equations (iv) and (v), we get
10 + 2e = b + c
10 + 2e = 28 [From (i)]
2e = 18

⇒ e = 9
4 + 9 = b
⇒ b = 13
9 + 6 = c
⇒ c = 15
Putting c = 15 in equation (ii), we get
d = 22 – 15 = 7
⇒ d = 7
Putting d = 7 in equation (vi), we get
6 + f = 7
⇒ f = 1∴ a = 28, b = 13, c = 15, d = 7, e = 9 and f = 1.

(ii) Let a, b, c, d, e, and f be the missing numbers.

40 + b = a …(i)
c + d = 40 …(ii)
d + 9 = b …(iii)
5 + e = c …(iv)
e + 7 = d …(v)
7 + f = 9
⇒ f = 2
Adding equations (iv) and (v), we get
2e + 12 = c + d
⇒ 2e + 12 = 40 [From (ii)]
⇒ 2e = 28
⇒ e = 14
c = 5 + 14 = 19
⇒ c = 19 [From (iv)]
14 + 7 = d
⇒ d = 21
b = 21 + 9 = 30
⇒ b = 30
a = 40 + 30 = 70
⇒ a = 70

∴ a = 70, b = 30, c = 19, d = 21, e = 14, and f = 2.

(iii) Let a, b, c, d, e, and f be the missing numbers.

a + b = 35 …(i)
c + d = 40 …(ii)
d + 7 = b …(iii)
Adding equations (ii) and (iii), we get
c + d + d +7 = a + b
⇒ c + 2d = 35 – 7 = 28 [From (i)]
⇒ c + 2d = 28 …(iv)
Also, 3 + 5 = c
⇒ c = 8
Putting c = 8 in equation (iv), we get
8 + 2d = 28
⇒ 2d = 28 – 8 = 2o
⇒ d = 10
8 + 10 = a
⇒ a = 18
10 + 7 = b
⇒ b = 17
5 + e = 10
⇒ e = 10 – 5 = 5
⇒ e = 5

5 + f = 7
⇒ f = 7 – 5 = 2
⇒ f = 2
∴ a = 18, b = 17, c = 8, d = 10, e = 5, and f = 2.

Figure it Out (Page 140)

Q1: Without building the entire pyramid, find the number in the topmost row given the bottom row in each of these cases.

(i) Bottom row: 4 , 13 , 8

Solution: We know that,

(a) Given, bottom row:

The number in the topmost row = 4 + 2 × 13 + 8
= 4 + 26 + 8
= 38

(b) Given, bottom row:

The number in the topmost row = 7 + 2 × 11 + 3
= 7 + 22 + 3
= 32

(c) Given, bottom row:The number in the topmost row = 10 + 2 × 14 + 25
= 10 + 28 + 25
= 63

Q2: Write an expression for the topmost row of a pyramid with 4 rows in terms of the values in the bottom row.

Ans: Let a, b, c, d, e, f, g, h, i, and j be the elements of the pyramid.

∴ e = a + b, f = b + c, g = c + d
h = e + f = (a + b) + (b + c) = a + 2b + c
i = f + g = (b + c) + (c + d) = b + 2c + d
j = h + i = (a + 2b + c) + (b + 2c + d) = a + 3b + 3c + d

Thus, the expression for the top row is (a + 3b + 3c + d).

Q3: Without building the entire pyramid, find the number in the topmost row given the bottom row in each of these cases.

(a) Bottom row: 8, 19, 21, 13

Ans: If a, b, c, and d are the bottom row, then the expression of the topmost row of the pyramid is a + 3b + 3c + d.
Given, bottom row;Here, a = 8, b = 19, c = 21, and d = 13
∴ The number in the topmost row = a + 3b + 3c + d
= 8 + 3(19) + 3(21) + 13
= 8 + 57 + 63 + 13
= 141
Thus, the number in the topmost row is 141.

(b) Bottom row: 7, 18, 19, 6

Ans: Given, bottom row:Here, a = 7, b = 18, c = 19 and d = 6
∴ The number in the topmost row = a + 3b + 3c + d
= 7 + 3(18) + 3(19) + 6
= 7 + 54 + 57 + 6
= 124
Thus, the number in the topmost row is 124.

(c) Bottom row: 9, 7, 5, 11

Ans: Given, bottom row:Here, a = 9, b = 1,c = 5, and d = 11
∴ The number in the topmost row = a + 3b + 3c + d
= 9 + 3(7)+ 3(5) + 11
= 9 + 21 + 15 + 11
= 56
Thus, the number in the topmost row is 56.

Q4: If the first three Virahāṅka-Fibonacci numbers are written in the bottom row of a number pyramid with three rows, fill in the rest of the pyramid. What numbers appear in the grid? What is the number at the top? Are they all Virahāṅka-Fibonacci numbers?

Ans: We know that the first three Virahanka-Fibonacci number sequence = 1, 2, 3
Here, the bottom rowLet a, b, and c be the missing numbers.b = 1 + 2 = 3
c = 2 + 3 = 5
and a = b + c = 3 + 5 = 8
The complete pyramid is:The numbers appear in the grid = 1, 2, 3, 3, 5, 8
∴ The number at the top = 8
Yes, 1, 2, 3, 3, 5, 8 are Virahanka-Fibonacci numbers.

Q5: What can you say about the numbers in the pyramid and the number at the top in the following cases?

(i) The first four Virahāṅka-Fibonacci numbers are written in the bottom row of a four row pyramid.

Ans: (i) We know that,
The first four Virahanka-Fibonacci numbers = 1, 2, 3, 5
Here, the bottom rowLet a, b, c, d, e, and f be the missing numbers.d = 1 + 2 = 3
e = 2 + 3 = 5
f = 3 + 5 = 8
b = d + e = 3 + 5 = 8
c = e + f = 5 + 8 = 13
and a = b + c = 8 + 13 = 21The numbers in the pyramid are 1, 2, 3, 5, 8, 8, 13, 21,….
We can say that the numbers are a Virahanka-Fibonacci sequence.
∴ The number at the top = 21

(ii) The first 29 Virahāṅka-Fibonacci numbers are written in the bottom row of a 29 row pyramid.

Ans: (ii) From the above solution, we get
The number at the top = 2 × (Total number of digits present at the bottom) – 1
= 2x – 1
= 2 × 29 – 1
= 58 – 1
= 57th
Fibonacci numbers.

Q6: If the bottom row of an n row pyramid contains the first n Virahāṅka-Fibonacci numbers, what can we say about the numbers in the pyramid? What can we say about the number at the top?

Ans: When the bottom row uses the first n Virahanka-Fibonacci numbers = (2n – 1)th
∴ The number at the top of the pyramid = (2n – 1)th Virahanka-Fibonacci number.

Page No. 142

Math Talk

Create your own calendar trick. For instance, choose a grid of a different size and shape.

Ans: (i) Add the 5 numbers in this grid and tell the sum.1 + 7 + 8 + 9 + 15 = 40
Let ‘a’ represent the topmost number.My own calendar trick.
Sum = a + (a + 6) + (a + 7) + (a + 8) + (a + 14) = 5a + 35.
Consider a 3 × 3 grid. Add the 9 numbers.(10 + 11 + 12) + (17 + 18 + 19) + (24 + 25 + 26) = 33 + 54 + 75 = 162
Let ‘a’ represent the top left number.My own calendar trick.
Sum = a + (a + 1) + (a + 2) + (a + 7) + (a + 8) + (a + 9) + (a + 14) + (a + 15) + (a + 16)
= 9a + 72
= 9(a + 8)
Consider a 1 × 3 grid.
Add the 3 numbers.28 + 29 + 30 = 87
Let ‘a’ represent the left number.My own calendar trick.
Sum = a + (a + 1) + (a + 2)
= 3a + 3
= 3(a + 1)

(ii)Let ‘a’ represent the topmost number.My own trick.Let ‘a’ represent the top left number.This trick works.Let ‘a’ be the top-most number.This trick works.

Q: In the following grids, find the values of the shapes and fill in the empty squares:

Ans: 

Page No. 144

Figure it Out

Q1: Fill the digits 1, 3, and 7 in _ _ × _ to make the largest product possible.

Ans: There are six ways to place three digits:
We can fill the first box with 1, 3, or 7.
For each of these choices, we have 2 ways of filling the remaining 2 digits.
The six choices are:

In each pair, the one with the larger multiplicand generates the larger product, so we can reduce the comparison to these three expressions.

Thus, the first term in both expressions is equal.

The second term shows that  it is the largest.

​Q2. Fill in the digits 3, 5, and 9 in  to make the largest product possible.

Ans: There are six ways to place three digits:
We can fill the first box with 3, 5, or 9.
For each of these choices, we have 2 ways of filling the remaining 2 digits.
The six choices are:

​In each pair, the one with the larger multiplicand generates the larger product, so we can reduce the comparison to these three expressions.

Thus, the first term in both expressions is equal.
The second term shows that  it is the largest.

Page 145-147

Figure It Out

Q1: In the trick given above, what is the quotient when you divide by 9? Is there a relationship between the two numbers and the quotient?

Ans: Let ab be the two-digit number. (b > a)
∴ ba > ab
The difference is (10b + a) – (10a + b) = 10b + a – 10a – b
= 9b – 9a
= 9(b – a), is divisible by 9.
When 9(b – a) is divided by 9, then the quotient is (b – a).

Q2: In the trick given above, instead of finding the difference of the two 2-digit numbers, find their sum. What will happen? 

For example:

• We start with 31. After reversing, we get 13. Adding 31 and 13, we get 44.
• We start with 28. After reversing, we get 82. Adding 28 and 82, we get 110.
• We start with 12. After reversing, we get 21. Adding 12 and 21, we get 33.

Observe that all these numbers are divisible by 11. Is this always true? Can we justify this claim using algebra?
Ans: 44, 110, 33 are divisible by 11.
Yes, it is always true.
44 = 4 – 4 = 0, divisible by 11.
110 = (1 + 0) – 1 = 0, divisible by 11.
33 = 3 – 3 = 0, divisible by 11.
Using Algebra
Original number = 10a + b
Reversed number = 10b + a
Sum = 10a + b + 10b + a = 11(a + b)

​Hence, the sum is always divisible by 11.

Math Talk

Q3: Consider any 3-digit number, say abc (100a + 10b + c). Make two other 3-digit numbers from these digits by cycling these digits around, yielding bca and cab. Now add the three numbers. Using algebra, justify that the sum is always divisible by 37. Will it also always be divisible by 3? [Hint: Look at some multiples of 37.]

Ans: abc = 100a + 10b + c
bca = 100b + 10c + a
cab = 100c + 10a + b
Sum of abc + bca + cab = 111a + 111b + 111c
= 111(a + b + c)
= 37 × 3(a + b + c), is always divisible by 37.
111 = 1 + 1 + 1 = 3, is always divisible by 3.
For example:
​Consider a number 153.
Other two numbers = 531 and 315
Sum = 153 + 531 + 315 = 999
999 = 37 × 27, which is divisible by 37.
999 = 9 + 9 + 9 = 27, which is also divisible by 3.

Math Talk

Q4: Consider any 3-digit number, say abc. Make it a 6-digit number by repeating the digits, that is abcabc. Divide this number by 7, then by 11, and finally by 13. What do you get? Try this with other numbers. Figure out why it works. [Hint: Multiply 7, 11 and 13.]

Ans: Given that abc is a 3-digit number.
abc = 100a + 10b + c
Make it a 6-digit number = abcabc
= 100000a + 10000b + 1000c + 100a + 10b + c
= 100100a + 10010b + 1001c
= 1001(100a + 10b + c)
The smallest number, divisible by 7, 11, and 13 = LCM (7, 11, 13)
= 7 × 11 × 13
= 1001
∴ abcabc = 1001(100a + 10b + c), is divisible by 7, 11, and 13.
Consider 836 a 3-digit number.
Make it 6-digit number = 836836 = 1001 × 836

​∴ 836836 is divisible by 7, 11, and 13.
This works because 10001 = 7 × 11 × 13, and repeating a 3-digit number creates a multiple of 1001.

Math Talk

Q5: There are 3 shrines, each with a magical pond in the front. If anyone dips flowers into these magical ponds, the number of flowers doubles. A person has some flowers. He dips them all in the first pond and then places some flowers in shrine 1. Next, he dips the remaining flowers in the second pond and places some flowers in shrine 2. Finally, he dips the remaining flowers in the third pond and then places them all in shrine 3. If he placed an equal number of flowers in each shrine, how many flowers did he start with? How many flowers did he place in each shrine?

Ans: Let x be the initial number of flowers, and k be the equal number of flowers placed in each of the three shrines.
In shrine 1, the remaining flowers = 2x – k
In shrine 2, the remaining flowers = 2(2x – k) – k
= 4x – 2k – k
= 4x – 3k
In shrine 3, the remaining flowers = 2(4x – 3k) – k
= 8x – 6k – k
= 8x – 7k
∴ 8x – 7k = 0
⇒ 8x = 7k
⇒ x = 7k/8
For the minimum possible number of flowers, we use the smallest positive integer k, which is k = 8.
∴ x = 7×8/8 = 7
Thus, the person started with 7 flowers and placed 8 flowers in each shrine.

Math Talk

Q6: A farm has some horses and hens. The total number of heads of these animals is 55 and the total number of legs is 150. How many horses and how many hens are on the farm? [Hint: If all the 55 animals were hens, then how many legs would there be? Using the difference between this number and 150, can you find the number of horses?]

Ans: Method 1: Using Algebra
Let x and y be the number of horses and hens, respectively.
According to the questions,
x + y = 55 ……..(i)
And, 4x + 2y = 150
⇒ 2x + y = 75 ……(ii)
Subtracting (i) from (ii), we get
2x + y – x – y = 75 – 55
⇒ x = 20
Putting x = 20 in equation (i), we get
20 + y = 55
⇒ y = 55 – 20 = 35
Thus, the number of horses = 20 and the number of hens = 35.
Method 2: (without letter numbers)
If all 55 animals were hens
Total legs would be 55 × 2 = 110 legs
But actual legs = 150
Difference = 150 – 110 = 40 legs
Each time we replace a hen with a horse.
We remove 2 legs (hen) and add 4 legs (horse).
Net increase = 2 legs
Number of horses needed = 40 ÷ 2 = 20
Number of hens = 55 – 20 = 35

Q7: A mother is 5 times her daughter’s age. In 6 years’ time, the mother will be 3 times her daughter’s age. How old is the daughter now?

Ans: Let the present age of the daughter = x years
And the present age of her mother = y years
According to the question,
5(x) = y
⇒ 5x = y …..(i)
In 6 years,
3(x + 6) = y + 6
⇒ 3x + 18 = y + 6
⇒ 3x + 18 – 6 = y
⇒ 3x + 12 = y …..(ii)
From equations (i) and (ii), we get
3x + 12 = 5x
⇒ 5x – 3x = 12
⇒ 2x = 12
⇒ x = 6
The present age of the daughter = 6 years

Q8: Two friends, Gauri and Naina, are cowherds. One day, they pass each other on the road with their cows. Gauri says to Naina, “You have twice as many cows as I do”. Naina says, “That’s true, but if I gave you three of my cows, we would each have the same number of cows”. How many cows do Gauri and Naina have?

Ans: Let x, y be the number of cows of Gauri and Naina.
According to the question,
2x = y …(i)
Also, x + 3 = y – 3
x – y = -3 – 3 = -6 ……(ii)
From equations (i) and (ii), we get
x – 2x = -6
⇒ -x = -6
⇒ x = 6
Putting x = 6 in equation (i), we get,
y = 2 × 6 = 12
Thus, Gauri and Naina have 6 and 12 cows, respectively.

Q9: I run a small dosa cart and my expenses are as follows:• Rent for the dosa cart is ₹5000 per day.• The cost of making one dosa (including all the ingredients and fuel) is ₹10.

(i) If I can sell 100 dosas a day, what should be the selling price of my dosa to make a profit of ₹2000?

Ans: Given, rent for the dosa cart = ₹ 5000/day.
The total cost of making one dosa = ₹ 10
(i) Given,
Number of dosas = 100
∴ The cost of making 100 dosas = 100 × ₹ 10 = ₹ 1000
Total cost price = Rent for the dosa cart + The cost of making 100 dosas
= ₹ 5000 + ₹ 1000
= ₹ 6000
Profit = ₹ 2000
∴ Total selling price = ₹ 6000 + ₹ 2000 = ₹ 8000
The selling price of one dosa = 8000/100 = ₹ 80

​(ii) Let n be the number of dosa.
Then total cost price = n × ₹ 10 + ₹ 5000
Total selling price = n × ₹ 50
Profit = ₹ 2000
S.P = C.P + profit
⇒ 50n = 10n + 5000 + 2000
⇒ 50n – 10n = 5000 + 2000
⇒ 40n = 7000
⇒ n = 7000/40
⇒ n = 175
So, I should sell 175 dosa.

Q10: Evaluate the following sequence of fractions:

$\frac{1}{3}$, (1+3)/(5+7), (1+3+5)/(7+9+11)

What do you observe? Can you explain why this happens? [Hint: Recall what you know about the sum of the first n odd numbers.]

Ans: 

Thus, the given sequences are equivalent fractions.
We know that the sum of the first n odd numbers is n2.
Numerators:
1 = 12 = 1
1 + 3 = 22 = 4
1 + 3 + 5 = 32 = 9
Denominators:
3 = 3 × 12
5 + 7 = 12 = 3 × 22
7 + 9 + 11 = 27 = 3 × 32
Thus, each fraction is 

Page No. 147

Q: Karim and the Genie

(i) How many coins did Karim initially have?

Ans: Let Karim have n coins initially
No. of coins after 1st round = 2n – 8
No. of coins after 2nd round = [2(2n – 8)] – 8
= 4n – 16 – 8
= 4n – 24
No. of coins after 3rd round = 2(4n – 24) – 8 = 0
⇒ 2(4n – 24) – 8 = 0
⇒ 8n – 48 – 8 = 0
⇒ 8n = 56
⇒ n = 7

​So, Karim initially had 7 coins.

​(ii) For what cost per round should Karim agree to the deal, if he wants to increase the number of coins he has?

Ans: Let c = cost per round (coins to give genie)
Starting with 7 coins:
After round 1: 2(7) – c = 14 – c
After round 2: 2(14 – c) – c = 28 – 3c
After round 3: 2(28 – 3c) – c = 56 – 7c
For Karim to increase his coins:
56 – 7c > 7
⇒ 56 – 7 > 7c
⇒ 49 > 7c
⇒ c < 7
The cost per round should be less than 7 coins.
For example, if c = 6:
After 3 rounds: 56 – 7(6) = 56 – 42 = 14 coins (doubled his money!)

​(iii) Through its magical powers, the genie knows the number of coins that Karim has. How should the genie set the cost per round so that it gets all of Karim’s coins?

Ans: Let Karim start with n coins, and let c = cost per round.
After 3 rounds, Karim has: 8n – 7c coins
For the genie to get all coins:
8n – 7c = 0
⇒ 7c = 8n
⇒ c = 8n/7
The genie should charge (8n)/7 coins per round, where n is Karim’s starting amount.
For this to be a whole number, n must be a multiple of 7.

NCERT Solutions: Tales by Dots and Lines

Page No. 103

Q: Calculate and mark the mean of each collection of data below.

Ans:

(a) Collection: 5, 7, 9

Mean = (5 + 7 + 9) ÷ 3 = 21 ÷ 3 = 7

The mean is 7.

(b) Collection: 2, 4, 6, 8

Mean = (2 + 4 + 6 + 8) ÷ 4 = 20 ÷ 4 = 5

The mean is 5.

(c) Collection: 10, 10, 11, 17

Mean = (10 + 10 + 11 + 17) ÷ 4 = 48 ÷ 4 = 12

The mean is 12.

Page No. 104

Math Talk

Q: Can you explain how the mean is the centre of each collection?

Ans: The mean is the centre because the sum of distances of all points to the left of the mean equals the sum of distances of all points to the right of the mean.

For example, in collection (c): 10, 10, 11, 17 with mean = 12:

• Distance on left: (12 – 10) + (12 – 10) + (12 – 11) = 2 + 2 + 1 = 5
• Distance on right: (17 – 12) = 5

Both distances are equal (5 units each), showing the mean balances the data.

Q: Mark the mean for the collections below.

Ans: (a) Collection: 3, 5, 7, 9, 11

Mean = (3 + 5 + 7 + 9 + 11) ÷ 5 = 35 ÷ 5 = 7

(b) Collection: 1, 3, 5, 7

Mean = (1 + 3 + 5 + 7) ÷ 4 = 16 ÷ 4 = 4

Math Talk

Q: Can you explain how the mean is the centre of each collection?

Ans: For collection (a): 3, 5, 7, 9, 11 with mean = 7:

• Distances on left: (7 – 3) + (7 – 5) = 4 + 2 = 6
• Distances on right: (9 – 7) + (11 – 7) = 2 + 4 = 6

For collection (b): 1, 3, 5, 7 with mean = 4:

• Distances on left: (4 – 1) + (4 – 3) = 3 + 1 = 4
• Distances on right: (5 – 4) + (7 – 4) = 1 + 3 = 4

In both cases, the total distances on both sides are equal, confirming the mean is the centre.

Q: Is the mean the midpoint of the two endpoints/extremes of the data?

Ans: No, the mean is not always the midpoint of the two extremes.

For example, in the collection 10, 10, 11, 17:

• Midpoint of extremes = (10 + 17) ÷ 2 = 27 ÷ 2 = 13.5
• Mean = 12

The mean (12) is different from the midpoint of extremes (13.5).

Instead, the mean is the point where the sum of distances on the left equals the sum of distances on the right.

Q: Can there be more than one such ‘centre’? In other words, is there any other value such that the sum of the distances to the values lower than it and the values higher than it will still be equal?

Ans: No, there cannot be more than one such centre.

Proof: Consider the collection 10, 10, 11, 17 with mean = 12.

If we take any value greater than 12:

• All distances on the left side will increase
• All distances on the right side will decrease
• The balance is broken

If we take any value less than 12:

• All distances on the left side will decrease
• All distances on the right side will increase
• Again, the balance is broken

Therefore, there is only one unique centre (the mean) where the sum of distances on both sides is equal.

Page No. 105

Q: Will including a new value in the data increase or decrease the mean?

Ans: It depends on the value being included:

1. If the new value is greater than the current mean: The mean will increase.

• Example: Data: 4, 6, 8 (mean = 6)
• Adding 10: New data: 4, 6, 8, 10
• New mean = (4 + 6 + 8 + 10) ÷ 4 = 28 ÷ 4 = 7
• The mean increased from 6 to 7.

2. If the new value is less than the current mean: The mean will decrease.

• Example: Data: 4, 6, 8 (mean = 6)
• Adding 2: New data: 2, 4, 6, 8
• New mean = (2 + 4 + 6 + 8) ÷ 4 = 20 ÷ 4 = 5
• The mean decreased from 6 to 5.

3. If the new value equals the mean: The mean will remain the same.

Math Talk

Q: What happens to the mean when an existing value is removed? When will the mean increase, decrease, or stay the same?

Ans: It depends on which value is removed:

1. If a value greater than the mean is removed: The mean will decrease.
   • Example: Data: 2, 4, 6, 10 (mean = 5.5)
   • Removing 10: New data: 2, 4, 6
   • New mean = (2 + 4 + 6) ÷ 3 = 12 ÷ 3 = 4
   • The mean decreased.
2. If a value less than the mean is removed: The mean will increase.
   • Example: Data: 2, 4, 6, 10 (mean = 5.5)
   • Removing 2: New data: 4, 6, 10
   • New mean = (4 + 6 + 10) ÷ 3 = 20 ÷ 3 = 6.67
   • The mean increased.
3. If a value equal to the mean is removed: The mean will stay the same.

Q: What happens to the mean if a value equal to the mean is included or removed?

Ans: Including a value equal to the mean: The mean remains unchanged.

• Example: Data: 3, 5, 7 (mean = 5)
• Adding 5: New data: 3, 5, 5, 7
• New mean = (3 + 5 + 5 + 7) ÷ 4 = 20 ÷ 4 = 5
• Mean stays 5.

Removing a value equal to the mean: The mean remains unchanged.

• Example: Data: 3, 5, 5, 7 (mean = 5)
• Removing 5: New data: 3, 5, 7
• New mean = (3 + 5 + 7) ÷ 3 = 15 ÷ 3 = 5
• Mean stays 5.

Fair-share interpretation: When we include or remove a value equal to the mean, we’re not changing the “fair share” amount. Each person already has their fair share, so adding or removing someone with exactly the fair share doesn’t change what everyone else has.

Page No. 106

Q: Explore if it is possible to include or remove 2 values such that the mean is unchanged.

Ans: Yes, it is possible. We need to include or remove two values whose average equals the current mean.

Example: Data: 4, 5, 6, 9 (mean = 6)

To keep the mean at 6, we can:

• Include 5 and 7 (their average = 6)
• New data: 4, 5, 5, 6, 7, 9
• New mean = (4 + 5 + 5 + 6 + 7 + 9) ÷ 6 = 36 ÷ 6 = 6

Or we can remove values:

• Remove 4 and 8 if 8 was in the data (their average = 6)
• This keeps the mean unchanged.

Q: How about including or removing 3 values without changing the mean? Is it possible?

Ans: Yes, it is possible. We need to include or remove three values whose average equals the current mean.

Example: Data: 3, 6, 9 (mean = 6)

To keep the mean at 6:

• Include 4, 6, and 8 (their average = 6)
• New data: 3, 4, 6, 6, 8, 9
• New mean = (3 + 4 + 6 + 6 + 8 + 9) ÷ 6 = 36 ÷ 6 = 6

Q: Can we include 2 values less than the mean and 1 value greater than the mean, so that the mean remains the same?

Ans: Yes, this is possible if the total of the three values equals 3 times the mean.

Example: Data: 5, 7, 9 (mean = 7)

To keep mean at 7:

• We need 3 values that total to $3 \times 7$ = 21
• Include 4, 6 (both less than 7) and 11 (greater than 7)
• Check: 4 + 6 + 11 = 21
• New data: 4, 5, 6, 7, 9, 11
• New mean = (4 + 5 + 6 + 7 + 9 + 11) ÷ 6 = 42 ÷ 6 = 7

Q: Try to include 2 values greater than the mean and 1 value less than the mean, so that the mean stays the same.

Ans:

Example: Data: 5, 7, 9 (mean = 7)

To keep mean at 7:

• We need 3 values that total to $3 \times 7$ = 21
• Include 8, 10 (both greater than 7) and 3 (less than 7)
• Check: 8 + 10 + 3 = 21
• New data: 3, 5, 7, 8, 9, 10
• New mean = (3 + 5 + 7 + 8 + 9 + 10) ÷ 6 = 42 ÷ 6 = 7

Relatively Unchanged!

Q: Consider the data: 8, 3, 10, 13, 4, 6, 7, 7, 8, 8, 5. Calculate its mean.

Ans: Sum = 8 + 3 + 10 + 13 + 4 + 6 + 7 + 7 + 8 + 8 + 5 = 79

Number of values = 11

Mean = 79 ÷ 11 = 7.18 (approximately)

Q: Now, consider this data with every value increased by 10: 18, 13, 20, 23, 14, 16, 17, 17, 18, 18, 15. What is its mean? Is there a quicker way to find out? [Hint: Observe the following dot plots corresponding to the two data collections.]

Ans: Method 1 (Direct calculation):Sum = 18 + 13 + 20 + 23 + 14 + 16 + 17 + 17 + 18 + 18 + 15 = 189 Mean = 189 ÷ 11 = 17.18 (approximately)

Method 2 (Quicker way):Since every value increased by 10, the mean also increases by 10. New mean = Original mean + 10 = 7.18 + 10 = 17.18

This is much quicker!

Observation: The relative position of the mean stays the same. The entire data set shifts up by 10, so the mean also shifts up by 10.

Page No. 107Try This

Q: Try to explain, using algebra, what the average is when a fixed number, e.g., 2 is subtracted from every value in the collection.

Ans: Let there be n values: x₁, x₂, x₃, … xₙ

Original average = (x₁ + x₂ + x₃ + … + xₙ) ÷ n = a

When 2 is subtracted from every value:

New average = [(x₁ – 2) + (x₂ – 2) + (x₃ – 2) + … + (xₙ – 2)] ÷ n

= [x₁ + x₂ + x₃ + … + xₙ – 2n] ÷ n

= [(x₁ + x₂ + x₃ + … + xₙ) ÷ n] – (2n ÷ n)

= a – 2

Therefore, the new average is 2 less than the original average.

Q: Try to explain this using the fair-share interpretation of average that you learnt last year.

Ans: Do it Yourself.

Q: What happens to the average if every value in the collection is doubled?

Ans: The average also doubles.

Example: Consider data: 3, 5, 7

Original mean = (3 + 5 + 7) ÷ 3 = 15 ÷ 3 = 5

Doubled data: 6, 10, 14

New mean = (6 + 10 + 14) ÷ 3 = 30 ÷ 3 = 10

New mean = 2 × Original mean

Algebraic Proof:

Let the n values be x₁, x₂, x₃, … xₙ with average a.

Original average: (x₁ + x₂ + x₃ + … + xₙ) ÷ n = a

When every value is multiplied by 5:

New average = [(5x₁) + (5x₂) + (5x₃) + … + (5xₙ)] ÷ n

= [5(x₁ + x₂ + x₃ + … + xₙ)] ÷ n (using distributive property)

= 5 × [(x₁ + x₂ + x₃ + … + xₙ) ÷ n]

= 5 × a

= 5a

Therefore, when every value is multiplied by 5, the new average is 5 times the original average.

Similarly, when every value is doubled, the average also doubles.

Page No. 113

Figure it Out

Math Talk

Q. Find the mean of the following data and share your observations:

(i) The first 50 natural numbers.

Ans: The first 50 natural numbers are: 1, 2, 3, 4, …, 50

Sum of first n natural numbers = n(n + 1) ÷ 2

Sum = $50 \times 51$ ÷ 2 = 2550 ÷ 2 = 1275

Mean = 1275 ÷ 50 = 25.5

Observation: The mean of the first n natural numbers is always (n + 1) ÷ 2. For 50 numbers, it’s 51 ÷ 2 = 25.5.

(ii) The first 50 odd numbers.

Ans: The first 50 odd numbers are: 1, 3, 5, 7, …, 99

Sum of first n odd numbers = n²

Sum = 50² = 2500

Mean = 2500 ÷ 50 = 50

Observation: The mean of the first n odd numbers is always n. For 50 odd numbers, the mean is 50.

(iii) The first 50 multiples of 4.

Ans: The first 50 multiples of 4 are: 4, 8, 12, 16, …, 200

This is 4 times the first 50 natural numbers.

Sum = 4 × (sum of first 50 natural numbers) = $4 \times 1275$ = 5100

Mean = 5100 ÷ 50 = 102

Observation: The mean of the first n multiples of 4 is 4 times the mean of the first n natural numbers. Mean = $4 \times 25$.5 = 102.

Q. The dot plot below shows a collection of data and its average; but one dot is missing. Mark the missing value so that the mean is 9 (as shown below).

Ans: From the dot plot, the data values
4, 7, 8, 8, 9, 9, 9, 9, 9, 11, x
Number of observations = 11
Mean = 9
Sum of data values = 4 + 7 + 8 + 8 + 9 + 9 + 9 + 9 + 9 + 11 = 83
Total sum = 9 × 11 = 99
Missing number = 99 – 83 = 16

Page No. 114

Q. Sudhakar, the class teacher, asks Shreyas to measure the heights of all 24 students in his class and calculate the average height. Shreyas informs the teacher that the average height is 150.2 cm. Sudhakar discovers that the students were wearing uniform shoes when the measurements were taken and the shoes add 1 cm to the height.

(i) Should the teacher get all the heights measured again without the shoes to find the correct average height? Or is there a simpler way?

Ans: (i) The teacher does not need to remeasure all the heights. Since the shoes add 1 cm to every student’s height, the average height with shoes is 1 cm more than the actual average height. So the correct average height is 1 cm less than the measured average height.

​(ii) What is the correct average height of the class?
(ii) Correct average height = 150.2 cm – 1 cm = 149.2 cm
∴ Option (d) is correct.

Q. The three dot plots below show the lengths, in minutes, of songs of different albums. Which of these has a mean of 5.57 minutes? Explain how you arrived at the answer.

Ans: We can determine it by examining the dot plots. In dot plot A, most song lengths are between 5 and 6.5 minutes, so the mean is likely around 5.57 minutes. In dot plots B and C, all song lengths are below 5.57 minutes, so their means cannot exceed 5.57. Therefore, dot plot A is the one with a mean of 5.57 minutes.
Check:
For Album A
Data values = 5, 5, 5.25, 5.5, 5.75, 6, 6.5

= 5.57 minutes
For Album B
Data values = 0.5, 0.75, 1.5, 1.5, 2, 3.75, 4.25, 5

= 2.41 minutes
For Album C
Data values = 3.5, 3.5, 3.5, 4, 4, 4, 4.25, 4.5

= 3.9 minutes
Album (A) has a mean of 5.57 minutes.

Math Talk

Q. Find the median of 8, 10, 19, 23, 26, 34, 40, 41, 41, 48, 51, 55, 70, 84, 91, 92.

Ans: The data is already in ascending order.

Number of values = 16 (even)

Median = Average of 8th and 9th values

8th value = 41, 9th value = 41

Median = (41 + 41) ÷ 2 = 82 ÷ 2 = 41

(i) If we include one value to the data (in the given list) without affecting the median, what could that value be?

Ans: To keep the median at 41, the new value should be 41 itself.

When we add 41, the data becomes: 8, 10, 19, 23, 26, 34, 40, 41, 41, 41, 48, 51, 55, 70, 84, 91, 92 (17 values)

The 9th value (middle value) is still 41.

Median remains 41

(ii) If we include two values to the data without affecting the median, what could the two values be?

Ans: With n = 18 (even) the median is the average of the 9th and 10th values.
To keep the median 41, we need to add two numbers whose sum is 82.
For this, one value should be less than 41 and the other greater than 41.
For example, the two values could be 40 and 42.

(iii) If we remove one value from the data without affecting the median, what could the value be?

Ans: With n = 15 (odd) the median is the 8th value that is 41.
So we can remove another 41 from the ordered list.

Page No. 115

Q. Examine the statements below and justify if the statement is always true, sometimes true, or never true.

(i) Removing a value less than the median will decrease the median.
​(ii) Including a value less than the mean will decrease the mean.
​(iii) Including any 4 values will not affect the median.
​(iv) Including 4 values less than the median will increase the median.

Ans: (i) Sometimes true: Removing a value less than the median can decrease the median. But if the data set has an even number of elements and the removed value is below the lower of the two middle values, the median may stay the same.
(ii) Always true: The mean is the sum of all values divided by the number of values. Adding a value less than the mean decreases the total sum less than proportionally to the increase in the number of values, thus decreasing the mean.
(iii) Sometimes true: Including any 4 values could shift the median depending on whether those values are above or below the median and on the original number of observations.
(iv) Never true: Including values less than the median will either keep the median the same or decrease it, but it will never increase it.

Q. The mean of the numbers 8, 13, 10, 4, 5, 20, y, 10 is 10.375. Find the value of y.

Ans: Mean = Sum of all values ÷ Number of values

10.375 = (8 + 13 + 10 + 4 + 5 + 20 + y + 10) ÷ 8

10.375 = (70 + y) ÷ 8

Multiplying both sides by 8:

83 = 70 + y

y = 83 – 70 = 13

The value of y is 13.

Q8. The mean of a set of data with 15 values is 134. Find the sum of the data.

Ans: Mean = Sum of data ÷ Number of values

134 = Sum ÷ 15

Sum = $134 \times 15$ = 2010

The sum of the data is 2010.

Q. Consider the data: 12, 47, 8, 73, 18, 35, 39, 8, 29, 25, p. Which of the following number(s) could be p if the median of this data is 29?

(i) 10  (ii) 25  (iii) 40  (iv) 100  (v) 29  (vi) 47  (vii) 30

Ans: Arranging the data (excluding p)
8, 8, 12, 18, 25, 29, 35, 39, 47, 73
Total observations = 11 (odd)
Median = (n+1/2)th term
= 6th term
= 29
So, p must be ≥ 29 to keep the value 29 in the middle.
So possible values of p are (iii) 40, (iv) 100, (v) 29, (vi) 47, and (vii) 30.

Q. The number of times students rode their cycles in a week is shown in the dot plot below. Four students rode their cycles twice in that week.

(i) Find the average number of times students rode their cycles.

Ans: 

Sum = (0 × 3) + (1 × 1) + (2 × 4) + (3 × 7) + (4 × 7) + (5 × 5) + (6 × 4) + (7 × 6) + (8 × 3) + (10 × 2)

= 0 + 1 + 8 + 21 + 28 + 25 + 24 + 42 + 24 + 20

= 193

Total number of students = 42

Average = 193/42 = 4.59 times

(ii) Find the median number of times students rode their cycles.

Ans: (ii) Total number of students = 42 (even)
So Median = average of 21st and 22nd terms

​The 21st and 22nd terms lie in the cumulative frequency corresponding to 4.
So median = 4+4/2 = 4
Median number of times students rode their cycles = 4

(iii) Which of the following statements are valid? Why?

(a) Everyone used their cycle at least once.
​(b) Almost everyone used their cycle a few times.
​(c) There are some students who cycled more than once on some days.
​(d) Exactly 5 students have used their cycles more than once on some days.
​(e) The following week, if all of them cycled 1 more time than they did the previous week, what would be the average and median of the next week’s data?

(iii) (a) Invalid
(b) Valid
(c) Valid
(d) Invalid
(e) Next week’s mean and median will be Mean = previous mean + 1
= 4.59 + 1
= 5.59
Median: Now the 21st and 22nd terms lie in the cumulative frequency corresponding to 5.
So Median = 5+5/2 = 5

Page No. 116

Q. A dart-throwing competition was organised in a school. The number of throws participants took to hit the bull’s eye (the centre circle) is given in the table below. Describe the data using its minimum, maximum, mean and median.

Given table:

Ans: The minimum number of trials is 1.
The maximum number of trials is 10.

Median:

Median = average of (62/2)th term and (62/2+1)th term
= average of 31th and 32th terms
The 31st and 32nd terms lie in the cumulative frequency corresponding to 8.

Page No. 122-123Figure it Out

Q: The average number of customers visiting a shop and the average number of customers actually purchasing items over different days of the week is shown in the table below. Visualise this data on a line graph.

Ans: 

Q2: The average number of days of rainfall in each month for a few cities is shown in the table below:

(i) What could be the possible method to compile this data?

Ans: Line graph

(ii) Mark the data for Mangaluru, Port Blair, and Rameswaram in the line graph shown below. You can round off the values to the nearest integer.

Ans: 

(iii) Based on the line for New Delhi in the graph, fill the data in the table.

Ans: 

(iv) Which city among these receives the most number of days of rainfall per year? Which city gets the least number of days of rainfall per year?

Ans: Mangaluru receives the most number of days of rainfall per year, and Rameswaram receives the least.

(v) Looking at the table, when is the rainy season in New Delhi and Rameswaram?

Ans: In New Delhi rainy season is from June to August, and in Rameswaram, it is from September to December.

Q: The following line graph shows the number of births in every month in India over a time period:

(i) What are your observations?
(ii) What was the approximate number of births in July 2017?
(iii) What time period does the graph capture?
(iv) Compare the number of births in the month of January in the years 2018, 2019, and 2020.
(v) Estimate the number of births in the year 2019.

Ans:  (i) There is both increasing and decreasing trends in the number of births in every month.
(ii) Approx 1.8 M.
(iii) The graph captures the time period from July 2017 to Jan 2020.
(iv) The number of births in Jan 2018 is approx 1.6 M, in Jan 2019 approx 1.8 M, and in Jan 2020 approx 1.9 M. That means it is increasing continuously.
(v) The number of births in the year 2019 is approx 1.9 M.Page No. 127Figure it Out

Q: Mean Grids:

(i) Fill the grid with 9 distinct numbers such that the average along each row, column, and diagonal is 10.

Ans: 

(ii) Can we fill the grid by changing a few numbers and still get 10 as the average in all directions?

Ans: Yes, we can fill by changing any number that requires adjusting other numbers to keep row, column, and diagonal sums equal to 3 × average = 3 × 10 = 30.

Q: Give two examples of data that satisfy each of the following conditions:

(i) 3 numbers whose mean is 8.

Ans: 3 numbers whose mean is 8 are
(a) 6, 8, and 10
(b) 7, 8, and 9.

(ii) 4 numbers whose median is 15.5.

Ans: 4 numbers whose median is 15.5 are
(a) 10, 15, 16, and 20
(b) 12, 15, 16, and 18.

(iii) 5 numbers whose mean is 13.6.

Ans: 5 numbers whose mean is 13.6 are
(а) 10, 12, 13, 15, and 18.
(b) 11, 12, 13, 14, and 18.

(iv) 6 numbers whose mean = median.

Ans: 6 numbers whose Mean = Median
(a) 2, 4, 6, 8, 10, 12 (Mean = Median = 7)
(b) 1, 3, 5, 7, 9, 11 (Mean = Median = 6)

(v) 6 numbers whose mean > median.

Ans: 6 numbers whose Mean > Median
(a) 1, 2, 3, 4, 5, 30 (Mean = 7.5, Median = 3.5)
(b) 2, 3, 4, 5, 6, 20 (Mean = 6.67, Median = 4.5)

Q: Fill in the blanks such that the median of the collection is 13: 5, 21, 14, _____, ______, ______. How many possibilities exist if only counting numbers are allowed?

Ans: The collection is (5, 21, 14, _____, ______, ______) with median 13.
For a set of 6 numbers,
the median (even) = average of (6/2)th and (6/2+1)th term
= average of 3rd and 4th terms
Let other terms be a, b, and c.
Arranging in order with a median of 13.
So, 14 must be the fourth term.
5, a, b, 14, 21, c
OR
5, a, b, 14 c, 21
6 + 14
Median = b+14/2 = 13
b = 26 – 14 = 12
∴ a < b and c > 14
One possible answer is 5, 8, 12, 14, 21, 28
If all three blanks are counting numbers and the median is 13, then infinite possibilities can exist as c > 14.

Q: Fill in the blanks such that the mean of the collection is 6.5: 3, 11, ____, _____, 15, 6. How many possibilities exist if only counting numbers are allowed?

Ans: The collection is (3, 11, x, y, 15, 6)
Mean = 6.5, let the other numbers be x and y.
For 6 numbers = 3+11+x+y+15+6/66 = 6.5
⇒ 35 + x + y = 6.5 × 6
⇒ 35 + x + y = 39
⇒ x + y = 39 – 35 = 4
With counting numbers, the pairs (x, y) satisfying x + y = 4 are (1, 3), (3, 1), and (2, 2).
So, the number of possibilities is 2.

Q: Check whether each of the statements below is true. Justify your reasoning. Use algebra, if necessary, to justify.

(i) The average of two even numbers is even.

Ans: True
Let the two even numbers be 2a and 2b.
Average = (2a+2b)/2 = a + b
Since (a + b) is an integer and the sum of two integers is an integer, the average is even.

(ii) The average of any two multiples of 5 will be a multiple of 5.

Ans: True
Let the two multiples of 5 be 5m and 5n.
Average = 5m+5n/2=5[(m+n)/2]
This is a multiple of 5 if [(m+n)/2] is an integer. That is m + n is even.

(iii) The average of any 5 multiples of 5 will also be a multiple of 5.

Ans: True
Let the five multiples of 5 be 5a, 5b, 5c, 5d, 5e.
Average = 5a+5b+5c+5d+5e/5
= 5[(a+b+c+d+e)/5]
= a + b + c + d + e
Since a + b + c + d + e is a whole number, the average of any 5 multiples of 5 will also be a multiple of 5.

Page No. 128

Q: There were 2 new admissions to Sudhakar’s class just a couple of days after the class average height was found to be 150.2 cm.

(i) Which of the following statements are correct? Why?

(a) The average height of the class will increase as there are 2 new values.
(b) The average height of the class will remain the same.
(c) The heights of the new students have to be measured to find out the new average height.
(d) The heights of everyone in the class has to be measured again to calculate the new average height.

Ans: (c) The heights of the new students have to be measured to find out the new average height.
The average may increase, decrease, or stay the same depending on the heights of the other 2 new students.

(ii) The heights of the two new joinees are 149 cm and 152 cm. Which of the following statements about the class’ average height are correct? Why?

(a) The average will remain the same.
(b) The average will increase.
(c) The average will decrease.
(d) The information is not sufficient to make a claim about the average.

Ans: (b) The average will increase.
Old mean = 150.2 cm
Let the number of students be x

​150.5 cm > 150.2 cm
As the average of 2 new added heights is greater than old average, the class average increases.
Option (b) is correct.

(iii) Which of the following statements about the new class average height are correct? Why?

(a) The median will remain the same.
(b) The median will increase.
(c) The median will decrease.
(d) The information is not sufficient to make a claim about median.

Ans: (d) The information is not sufficient to make a claim about the median.
The median depends on the arrangement of all heights in the ordered list.
Adding two values changes the position of the middle value.
Option (d) is correct.

Math Talk
Q: Is 17 the average of the data shown in the dot plot below? Share the method you used to answer this question.

Ans: Method: Count and Calculate

Sum = (14 × 2) + (15 × 2) + (16 × 3) + (17 × 5) + (18 × 4) + (19 × 4) + (20 × 3) + 21 + 23
= 28 + 30 + 48 + 85 + 72 + 76 + 60 + 21 + 23
= 443
Total numbers = 2 + 2 + 3 + 5 + 4 + 4 + 3 + 1 + 1 = 25
Mean = 443/25 = 17.72
The average of the data is 17.72, or on a dot plot, 17.

Math Talk

Q: The weights of people in a group were measured every month. The average weight for the previous month was 65.3 kg and the median weight was 67 kg. The data for this month showed that one person has lost 2 kg and two have gained 1 kg. What can we say about the change in mean weight and median weight this month?

Ans: Original average weight = 65.3 kg
Let there be n people in the group.
Total weight = n × 65.3 kg
After one person loses 2 kg and two people each gain 1 kg, the net change in total weight is
Total weight = 65.3n + 1 + 1 – 2 = 65.3n
New mean = 65.3n/n = 65.3 kg
The new mean weight will remain unchanged.
The effect on the median depends on the original position of these three people, whose weight gains or lose.
So new median weight cannot be determined exactly without knowing more data.

Page No. 129

Q: The following table shows the retail price (in ₹) of iodised salt in the month of January in a few states over 10 years.

(i) Choose data from any 3 states you find interesting and present it through a line graph using an appropriate scale.

Ans: (i) 

(ii) What do you find interesting in this data? Share your observations.

Ans: Price generally increases over the years for most states.
Mizoram shows the highest price jump and the highest overall prices in later years.
Gujarat has relatively stable prices compared to other states.

(iii) Compare the price variation in Gujarat and Uttar Pradesh.

Ans: Uttar Pradesh has a large price increase compared to Gujarat.

(iv) In which state has the price increased the most from 2016 to 2025?

Ans: Calculating the price increase for each state.
Andaman and Nicobar Islands = 20.99 – 16 = 4.99
Assam = 12.35 – 6 = 6.35
Gujarat = 19.2 – 16.5 = 2.7
Mizoram = 29.8 – 20 = 9.8
Uttar Pradesh = 24.81 – 16.15 = 8.66
West Bengal = 23.99 – 9.47 = 14.52
The price in West Bengal increases the most (₹ 14.42).

(v) What are you curious to explore further?

Ans: We want to explore the reason behind the sharp price rise in West Bengal.

Page No. 130 – 131

Q: Referring to the graph below, which of the following statements are valid? Why?

[Graph showing percentage of households using electricity and kerosene as primary lighting source in rural and urban areas from 1981 to 2023]

(i) In 1983, the majority in rural areas used kerosene as a primary lighting source while the majority in urban areas used electricity.

Ans: In 1983, approx 85% majority in rural areas used kerosene, whereas 65% majority in urban areas used electricity. So the statement is valid.

(ii) The use of kerosene as a primary lighting source has decreased over time in both rural and urban areas.

Ans: Valid statement as the graph shows a decreasing trend of kerosene in both areas.

(iii) In the year 2000, 10% of the urban households used electricity as a primary lighting source.

Ans: Invalid statement.
​The graph shows that more than 80% urban majority use electricity as a primary lighting source.

(iv) In 2023, there were no power cuts.

Ans: The graphs do not give any information about the power cut. So the statement is invalid.

Q: Answer the following questions based on the line graph.

(i) How long do children aged 10 in urban areas spend each day on hobbies and games?

Ans: (i) Children aged 10 years in urban areas spend approximately 2 hours each day on hobbies and games.

(ii) At what age is the average time spent daily on hobbies and games by rural kids 1.5 hours?

Options:(a) 8 years (b) 10 years (c) 12 years (d) 14 years (e) 18 years

Ans: (d) 14 years

(iii) Are the following statements correct?

(a) The average time spent daily on hobbies and games by kids aged 15 is twice that of kids aged 10.
​(b) All rural kids aged 15 spend at least 1 hour on hobbies and games everyday.

Ans: (a) and (b) both statements are correct.

Q: Individual project: Make your own activity strip for different days of the week.

(i) Do you eat and sleep at regular times every day? Typically how long do you spend outdoors?
(ii) Calculate the average time spent per activity. Represent this average day using a strip.
(iii) Similarly, track the activities of any adult at home. Compare your data with theirs.

Ans: Collect and analyse your own data to answer the question.

Q: Small group project: Make a group of 3 – 4 members. Do at least one of the following:

(i) Track daily sleep time of all your family members for a week.

(a) Represent this on strips.
(b) Put together the data of all your group members. Calculate the average and median sleep time of children, adults, elderly.
(c) Share your findings and observations.

(ii) When do schools start and end? Collect information on the daily timings of different schools for Grade 8, including class time and break time. Analyse and present the data collected.

Ans: Collect and analyse your own data to answer the question.

Page No. 131 – 132

Q14: The following graphs show the sunrise and sunset times across the year at 4 locations in India. Observe how the graphs are organised.

Are you able to identify which lines indicate the sunrise and which indicate the sunset?

Q: Answer the following questions based on the graphs:

(i) At which place does the sun rise the earliest in January? What is the approximate day length at this place in January?
​(ii) Which place has the longest day length over the year?
​(iii) Share your observations — what do you find interesting? What are you curious to find out?

Ans: Yes, we can identify the lines.
(i) In January, the sunrise was earliest in Kibithu. (approx 6:00)
The day length is (16:00-6:00) = 10 hours
(ii) The place with the longest day length over the year is Ghuar Moti.
(iii) Observations are
Longest day length – Ghuar Moti
Shortest day length – Kanyakumari
Earliest sunrise in Jan – Kibithu
Latest sunset in Jan – Ghuar Moti
I am curious to find out the seasonal variation of sunrise and sunset, or day length, for any particular location.

Q: We all know the typical sunrise and sunset timings. Do you know when the moon rises and sets? Does it follow a regular pattern like the sun? Let’s find out.

The following graph shows the moonrise and moonset times over a month.

(i) Find out on what dates amavasya (new moon) and purnima (full moon) were in this month.
(ii) What do you notice? What do you wonder?

Ans: (i) Amavasya – Day 21
Purnima – Day 7-8
(ii) Observation: The difference between moonrise and moonset changes throughout the month.
Around the middle of the month, the gap between moonrise and moonset is the maximum.
I wonder
Why do the moonrise and moonset times change every day?
Why is the time gap between moonrise and moonset maximum on some days and minimum on others?

NCERT Solutions: Exploring Some Geometric Themes

Page No. 72

Figure it Out

Q: Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Sierpinski Carpet.

Ans: Step 0: Take cut out of an equilateral triangle Δ.
Step 1: Divide it into 4 equilateral triangles by joining the midpoints of each of the sides.
Remove the central triangle.

Step 2: Divide each of the remaining 3 equilateral triangles into four equilateral triangles and remove the central triangle in each of them.

Step 3: Repeat the steps again and again to get Sierpinski’s gasket.

Q: Find the number of holes, and the triangles that remain at each step of the shape sequence that leads to the Sierpinski Triangle.

Ans: Number of holes in
Step 0: 0 hole
Step 1: 1 hole
Step 2: 1 + 3 = 4 holes
Step 3: 1 + 3 + 9 = 13 holes
Step 4: 1 + 3 + 9 + 27 = 40 holes
.
.
.

Q: Find the area of the region remaining at the nth step in each of the shape sequences that lead to the Sierpinski fractals. Take the area of the starting square/triangle to be 1 sq. unit.

Ans:

For Sierpinski Carpet:

Let An = area remaining at step n

• Step 0: A₀ = 1 sq. unit (full square)
• Step 1: A₁ = $\frac{8}{9}$ sq. unit (8 squares out of 9 remain)
• Step 2: A₂ = ($\frac{8}{9}$) × ($\frac{8}{9}$) = ($\frac{8}{9}$)² sq. unit

At each step, we keep $\frac{8}{9}$ of the area from the previous step.

Therefore, An = ($\frac{8}{9}$)ⁿ sq. unit

For Sierpinski Triangle:

Let An = area remaining at step n

• Step 0: A₀ = 1 sq. unit (full triangle)
• Step 1: A₁ = $\frac{3}{4}$ sq. unit (3 triangles out of 4 remain)
• Step 2: A₂ = ($\frac{3}{4}$) × ($\frac{3}{4}$) = ($\frac{3}{4}$)² sq. unit

At each step, we keep $\frac{3}{4}$ of the area from the previous step.

Therefore, An = ($\frac{3}{4}$)ⁿ sq. unit

Page No. 73Koch Snowflake

Figure it Out

Q: Draw the initial few steps (at least till Step 2) of the shape sequence that leads to the Koch Snowflake.

Ans: 

Q: Find the number of sides in the nth step of the shape sequence that leads to the Koch Snowflake.

Ans: Number of sides:
Step 0: 3
Step 1: 3 × 4 = 12
Step 2: 3 × 4² = 48
Step 3: 3 × 4³ = 192
.
.
.
Step n: 3 × 4ⁿ

Q: Find the perimeter of the shape at the nth step of the sequence. Take the starting equilateral triangle to have a sidelength of 1 unit.

Ans:

Let Pn = perimeter at step n

Step 0:

• Side length = 1 unit
• Number of sides = 3
• P₀ = $3 \times 1$ = 3 units

Step 1:

• Each side of length 1 is divided into 3 parts, each of length $\frac{1}{3}$
• Each side becomes 4 segments of length $\frac{1}{3}$
• Number of sides = 12
• P₁ = 12 × ($\frac{1}{3}$) = 4 units

Step 2:

• Each side of length $\frac{1}{3}$ becomes 4 segments of length $\frac{1}{9}$
• Number of sides = 48
• P₂ = 48 × ($\frac{1}{9}$) = $\frac{16}{3}$ units

Pattern:

• At step n, each original side is divided into 3ⁿ equal parts
• Length of each small segment = 1/3ⁿ
• Number of sides = $3 \times 4$ⁿ
• Perimeter Pn = ($3 \times 4$ⁿ) × (1/3ⁿ)

Therefore, Pn = 3 × ($\frac{4}{3}$)ⁿ units

Note: As n increases, the perimeter keeps increasing and approaches infinity!

Page No. 75Build it in Your Imagination

Q: Picture your name, then read off the letters backwards. Make sure to do this by sight, not by sound — really see your name! Now try with your friend’s name.

Ans: This is a visualization exercise. Let’s take an example:

If your name is “RAVI”:

• Visualize: R-A-V-I
• Reading backwards by sight: I-V-A-R

If your friend’s name is “PRIYA”:

• Visualize: P-R-I-Y-A
• Reading backwards by sight: A-Y-I-R-P

Practice this with different names to improve visual memory and spatial thinking.

Page No. 76

Q: Cut off the four corners of an imaginary square, with each cut going between midpoints of adjacent edges. What shape is left over? How can you reassemble the four corners to make another square?

Ans: ABCD is a square.
P, Q, R, and S are midpoints of AB, BC, CD, and AD, respectively.

When we cut along PQ, QR, RS, and PS, we again get a square.
If we join the shaded portions again, we get a square of the same size as that of PQRS.

Q: Mark the sides of an equilateral triangle into thirds. Cut off each corner of the triangle, as far as the marks. What shape do you get?

Ans: XYZ is an equilateral triangle with AY = YZ = ZX.
Divide each side into 3 equal parts as shown.

We note XA = AB = BY = YC = CD = DZ = ZE = EF = FX
Join AF, BC, ED.

Shape ABCDEF is a hexagon.
As all sides of the hexagon are equal, it is called regular hexagon.

Q: Mark the sides of a square into thirds and cut off each of its corners as far as the marks. What shape is left?

Ans: KLMN is a square.
Each side of the square is divided into three equal parts.

When the four corners of the square are cut off, we get an octagon.

Page No. 77

Q: A solid whose profile has a square outline

Ans: A solid with a profile that has a square outline is a CUBE.

Q: A solid whose profile has a circular outline

Ans: A solid whose profile has a circular outline is a sphere

Q: A solid whose profile has a triangular outline

Ans: A solid whose profile has a triangular outline is a triangular pyramid.

Q: A solid with a rectangular profile from one viewpoint and a circular profile from another viewpoint

Ans: 

Q: A solid with a circular profile from one viewpoint and a triangular one from another viewpoint

Ans: 

Q: A solid with a rectangular profile from one viewpoint and a triangular one from another viewpoint

Ans: 

Q: A solid with a trapezium shaped profile from one viewpoint and a circular one from another viewpoint

Ans: A truncated cone or a cone cut off from a larger cone.
This shape is called a frustum.

Q: A solid with a pentagonal profile from one viewpoint and a rectangular one from another viewpoint

Ans: Pentagon Base Prism

Page No. 79Math Talk Questions

Q: If the congruent polygons of a prism have 10 sides, how many faces, edges and vertices does the prism have? What if the polygons have n sides?

Ans: Number of sides of a congruent polygon of a prism = 10
Number of faces = n + 2
= 10 + 2
= 12
Number of edges = 3 × n
= 3 × 10
= 30
Number of vertices = 2 × n
= 2 × 10
= 20
A polygon with n sides will have
Number of faces = n + 2
Number of edges = 3n
Number of vertices = 2n

Q: If the base of a pyramid has 10 sides, how many faces, edges and vertices does the pyramid have? What if the base is an n-sided polygon?

Ans: Number of sides of the base of the pyramid = 10
Number of faces = n + 1
= 10 + 1
= 11
Number of edges = 2n
= 2 × 10
= 20
Number of vertices = n + 1
= 10 + 1
= 11
If the base is an n-sided polygon, then
No. of faces = n + 1
No. of edges = 2n
No. of vertices = n + 1

Page No. 80 – 81Figure it Out

Q: Which of the following are the nets of a cube? First, try to answer by visualisation. Then, you may use cutouts and try.

(i), (ii), (iii), (iv), (v), (vi)

Ans: (i) No
(ii) Yes
(iii) Yes
(iv) Yes
(v) No
(vi) Yes

Q: A cube has 11 possible net structures in total. In this count, two nets are considered the same if one can be obtained from the other by a rotation or a flip. Find all the 11 nets of a cube.

Ans: 

Q: Draw a net of a cuboid having sidelengths:

(i) 5 cm, 3 cm, and 1 cm
(ii) 6 cm, 3 cm, and 2 cm

Ans: 

Q: What is a net of a regular tetrahedron? Which of the following are nets of a regular tetrahedron? Are there any other possible nets?

Ans: (ii) and (iv) are not a net of a tetrahedron.

Math Talk

Q: Draw a net with appropriate measurements that can be folded into a regular tetrahedron. Verify if it works by making an actual cutout.

Ans: Draw an equilateral triangle of side length 6 cm.
Mark the midpoints of the sides of the triangle.
Join the midpoints. Fold along the dotted lines to get the tetrahedron.

Cut out the triangle to form an actual tetrahedron.

Q: Draw a net with appropriate measurements that can be folded into a square pyramid. Verify if it works by making an actual cutout.

Ans: (a) Draw four squares of side 4 cm in a row.

(b) On the opposite edges of any of the middle two squares, draw two more squares of side 4 cm each.

Fold along the common edges to form a cube of edge length 4 cm.
Cut along the outer boundary and fold along the common edge to form a cube.

Q: What is the net of a cylinder? 
Ans: A net of a cylinder is a rectangle and two equal circles.
For the net of a cylinder, draw a circular face of radius r cm on the opposite side of a rectangle with length = 2πr and breadth = h.

Page No. 82

Q: What are the sidelengths of the rectangle obtained (from unfolding a cylinder)?

Ans: When a cylinder is unfolded into its net:

The rectangle has:

• Length = Circumference of the circular base = 2πr (where r is the radius of the base)
• Width (or height) = h (the height of the cylinder)

Explanation:

• When we “unroll” the curved surface, the distance around the circle becomes the length of the rectangle
• The height of the cylinder becomes the width of the rectangle
• The two circular faces remain as circles

Complete net of cylinder consists of:

• 1 rectangle with dimensions 2πr × h
• 2 circles with radius r

Q: How will the net of a cone look?

Ans: To make a net of a cone, draw a sector of a circle of radius R with a length of arc of the sector as 2πr and a circle of radius r attached to the arc.

Math Talk

Q: What surface do you construct by using the above net, in which O is not the centre of the boundary circle? Make a physical model to help you answer this question!

Ans: When we join the radii of the sector, we get a cone:

​​OA and OB are radii of a circle.
Paste OA over OB, we get a cone with slant height = OA.

Q: Draw a net with appropriate measurements that can be folded into a triangular prism. Verify that it works by making an actual cutout.

Ans: ABCD is a rectangle of sides 6 cm by 2 cm. DCIF and FIHG are also rectangles of the same dimensions.
DEF and CJI are equilateral triangles of side 2 cm each.

​By folding along the common edges, we get a triangular prism.Page No. 92 – 93

Figure it Out

Q: Observe the front view, top view and side view of the different lines in Fig. 4.6. Is there any relation between their lengths?

Ans: (a) Horizontal Line
(b) oblique (slanting) line
(c) oblique (slanting) line, More tilted than line (b)
Top views show that (a) is the shortest, and (c) is the longest

Q: Find the front view, top view and side view of each of the following solids, fixing its orientation with respect to the vertical, horizontal and side planes: cube, cuboid, parallelepiped, cylinder, cone, prism, and pyramid.

Ans: (a) Cube: Assuming standard orientation with faces parallel to planes.

(ii) Cuboid (dimensions: length l, breadth b, height h)

(iii) Parallelepiped (all faces are parallelograms)

(iv) Cylinder (axis vertical)

(v) Cone (axis vertical, base on horizontal plane)

(vi) Prism (regular prism with square base, axis vertical)

(vii) Pyramid (square pyramid, axis vertical)

Q: Match each of the following objects with its projections.

Ans: (a) – (viii)
(b) – (vi)
(c) – (vii)
(d) – (i)
(e) – (iii)
(f) – (iv)
(g) – (v)
(h) – (ii)Page No. 95

Figure it Out

Q1: Draw the top view, front view and the side view of each of the following combinations of identical cubes.

Ans: Do it Yourself.Page No. 95

Math Talk

Q: Imagine eight identical cubes, glued together along faces to form the letter ‘E’.

(i) This looks like a ‘E’ from the front. What does it look like from the side? From the top?
​(ii) Glue additional cubes to make a shape that looks like ‘E’ from the front and ‘L’ from the top.
(iii) Now, can you glue even more cubes to make it look like ‘E’ from the front, ‘L’ from the top, and ‘F’ from the side?
(iv) Can you think of other letter combinations to make with a single combination of cubes in this manner?

Ans: Page No. 96

Q: Which solid corresponds to the given top view, front view, and side view?

Ans: Solid (ii)

Q: Using identical cubes, make a solid that gives the following projections.

Ans: 

Page No. 97

Q: Find the number of cubes in this stack of identical cubes.

Ans: Counting from the top layer to the bottom layer:
1 + 3 + 6 + 10 = 20 cubes

Q: What are the different shapes the projection of a cube can make under different orientations?

Ans: Five different shapes can be observed.
(a) Square
Orientation: One face of the cube is parallel to the projection plane.
(b) Rectangle
Orientation: Two faces are visible, but one set of edges is parallel to the plane.
(c) Parallelogram
Orientation: A face is tilted relative to the plane.
(d) Rhombus
Orientation: A special tilted case where all projected edges remain equal.
(e) Hexagon (maximum case)
Orientation: The cube is oriented so that three faces are equally visible (e.g., looking along a body diagonal).

Page No. 100

Figure it Out

Math Talk

Q: In addition to the 5 ways shown in Fig. 4.8, are there any additional ways of gluing four cubes together along faces? Can you visualise and draw these as well?

Ans: Page No. 101 – 102

Q: Draw the following figures on the isometric grid.

Hint: It may be useful to determine whether the edge to be currently drawn — say, along the height — goes from down to up or up to down. Accordingly, draw the line segment on the grid either in the direction of the height axis or opposite to it.

Ans: 

Q: Is there anything strange about the path of this ball? Recreate it on the isometric grid.

Hint: Consider a portion of this figure that is physically realisable and identify the 3 primary directions.

Ans: The picture shows the Penrose staircase. It is an optical illusion showing a loop of stairs that appears to rise or descend forever. Each step looks locally consistent, but the structure cannot exist in reality. The illusion works by exploiting perspective and depth cues, creating the impression of continuous motion without a true beginning or end. On a Penrose staircase, a ball would have no physically possible path at all—because the staircase itself cannot exist as a single, consistent object in real 3D space. However, we can still answer the question in two ways:

• Every step appears to slope downward, yet the staircase loops back to the starting point.
  When the ball is released, it rolls “down” the stairs, goes around the loop, and keeps rolling forever, always downhill. This creates a perpetual- motion illusion — a never-ending descent with no lowest point.
• If we have a Penrose staircase in the real world, at least one section would slope upward or the staircase would have to twist or break. The loop would not close. So the ball would roll down until it reaches a lowest point, then stop or roll back the way it came. There is no continuous path where the ball can always go down and still return to the start.

Q: Observe this triangle.

(i) Would it be possible to build a model out of actual cubes? What are the front, top, and side profiles of this impossible triangle?
​(ii) Recreate this on an isometric grid.
(iii) Why does the illusion work?

Ans: The impossible triangle using cubes creates an optical illusion where three straight beams of square cross-section appear to form a continuous, closed loop.
It can be built using cubes of the same size in the following steps.

​Step 1: Bottom Row: Lay a horizontal row of 4-5 cubes.
Step 2: Vertical Row: At one end, stack 4-5 cubes vertically to form a 90-degree corner.
Step 3: The “Gap” Row: At the top of the vertical stack, extend a row of cubes horizontally away from the viewer (into the depth of the scene)
When observed through a camera at a specific angle, the end of this third row will appear to “touch” the first horizontal row, even though they are feet apart.
Front, side, and top views are as follows.

(ii)

(iii) The impossible triangle works because:

• Our brain assumes a 3D structure from 2D images.
• Our brain focuses on small, local areas rather than checking the entire object for logical consistency. When we mentally “connect” the corners, the contradictions are hidden.
• Our brain doesn’t notice the switch—it keeps one depth interpretation and ignores the rest.
• Our visual system follows a principle called good continuation: it prefers smooth, continuous shapes rather than broken ones.
• So instead of seeing three separate bars that don’t line up, your brain forces them into a single triangle.

NCERT Solutions: Proportional Reasoning – 2

Page No. 60Figure it Out

Q1: A cricket coach schedules practice sessions that include different activities in a specific ratio — time for warm-up/cool-down : time for batting : time for bowling : time for fielding :: 3 : 4 : 3 : 5. If each session is 150 minutes long, how much time is spent on each activity?

Ans: Given, time for warm-up/cool-down : time for batting : Time for bowling : time for fielding :: 3 : 4 : 3 : 5

Total number of ratio parts = 3 + 4 + 3 + 5 = 15

Total time of each session = 150 minutes

So, time for warm-up/cool-down = $\frac{3}{15}\times 150=30$ minutes

Time for batting = $\frac{4}{15}\times 150=40$ minutes

Time for bowling = $\frac{3}{15}\times 150=30$ minutes

Time for fielding = $\frac{5}{15}\times 150=50$ minutes

Verification: 30 + 40 + 30 + 50 = 150 minutes

Q2: A school library has books in different languages in the following ratio — no. of Odiya books : no. of Hindi books : no. of English books :: 3 : 2 : 1. If the library has 288 Odiya books, how many Hindi and English books does it have?

Ans: Books Ratio

Given, No. of Odiya books : No. of Hindi books : No. of English books :: 3 : 2 : 1

Let x be the total number of books.

No. of Odiya books = $\frac{3}{6}\times x$

⇒ 288 = $\frac{3}{6}\times x$

⇒ x = 576

∴ No. of Hindi books = $\frac{2}{6}\times 576=192$

and no. of English books = $\frac{1}{6}\times 576=96$

Q3: I have 100 coins in the ratio — no. of ₹10 coins : no. of ₹5 coins : no. of ₹2 coins : no. of ₹1 coins :: 4 : 3 : 2 : 1. How much money do I have in coins?

Ans: Coins Ratio

Given no. of ₹ 10 coins : no. of ₹ 5 coins : no. of ₹ 2 coins : no. of ₹ 1 coins :: 4 : 3 : 2 : 1.

Total number of coins = 100

Total number of ratio parts = 4 + 3 + 2 + 1 = 10

no. of ₹ 10 coins = $\frac{4}{10}\times 100=40$

no. of ₹ 5 coins = $\frac{3}{10}\times 100=30$

no. of ₹ 2 coins = $\frac{3}{10}\times 100=30$

no. of ₹ 1 coins = $\frac{1}{10}\times 100=10$

Total money = 40 × 10 + 30 × 5 + 2 × 20 + 1 × 10

= 400 + 150 + 40 + 10

= ₹ 600

Math Talk

Q4: Construct a triangle with sidelengths in the ratio 3 : 4 : 5. Will all the triangles drawn with this ratio of sidelengths be congruent to each other? Why or why not?

Ans: We can construct triangles with sides in the ratio 3 : 4 : 5.
They will not be congruent to each other.
Reason:
Triangle 1: Let sides 3 cm, 4 cm, 5 cm
Triangle 2: Let sides 6 cm, 8 cm, 10 cm
Triangle 3: Let sides = 9 cm, 12 cm, 15 cm
Though all these triangles have the same ratio (3 : 4 : 5), their actual sizes are different.
Congruent triangles must have the same shape and size.
These triangles have the same shape (they are similar) but different sizes.
Hence, they are not congruent.

Q5: Can you construct a triangle with sidelengths in the ratio 1 : 3 : 5? Why or why not?

Ans: For a triangle to exist, it must satisfy the triangle inequality theorem, which states:
The sum of any two sides of a triangle must be greater than the third side.
Let’s take
Side 1 = 1 cm
Side 2 = 3 cm
Side 3 = 5 cm
1. 1 + 3 < 5 ⇒ 4 > 5 (No)
2. 1 + 5 > 3 ⇒ 6 > 3 (Yes)
3. 3 + 5 > 1 ⇒ 8 > 1 (Yes)
Since the first condition fails (1 + 3 = 4 < 5)
We can’t construct a triangle with these sidelengths.

Page No. 62-63Figure it Out

Q1: A group of 360 people were asked to vote for their favourite season from the three seasons — rainy, winter and summer. 90 liked the summer season, 120 liked the rainy season, and the rest liked the winter. Draw a pie chart to show this information.

Ans:  Given, total people = 360
90 people liked the summer season.
120 people liked the rainy season.

∴ People liked winter season = 360 − (120 + 90) = 150

So, angle for summer season = $\frac{90}{360}\times 360^{\circ }=90^{\circ }$

Angle for rainy season = $\frac{120}{360}\times 360^{\circ }=120^{\circ }$

Angle for winter season = $\frac{150}{360}\times 360^{\circ }=150^{\circ }$

Verification: 90 + 120 + 150 = 360°

Q2: Draw a pie chart based on the following information about viewers’ favourite type of TV channel: Entertainment — 50%, Sports — 25%, News — 15%, Information — 10%.

Ans: Given, Entertainment = 50%
Sports = 25%
News = 15%
Information = 10%

 Angle for entertainment = 50% of 360° 

= $\frac{50}{100}\times 360^{\circ }$

= 180°

Angle for sports = 25% of 360°

= $\frac{25}{100}\times 360^{\circ }$

= 90°

Angle for news = 15% of 360°

= $\frac{15}{100}\times 360^{\circ }$

= 54°

Angle for information = 10% of 360°

= $\frac{10}{100}\times 360^{\circ }$

= 36°

Verification: 180° + 90° + 54° + 36° = 360°

Q3: Prepare a pie chart that shows the favourite subjects of the students in your class.

Ans:

Total number of students = 4 + 6 + 9 + 3 + 10 + 12 + 16 = 60

Angle for Language = $\frac{4}{60}\times 360^{\circ }=24^{\circ }$

Angle for Arts Education = $\frac{6}{60}\times 360^{\circ }=36^{\circ }$

Angle for Vocational Education = $\frac{9}{60}\times 360^{\circ }=54^{\circ }$

Angle for Social Science = $\frac{3}{60}\times 360^{\circ }=18^{\circ }$

Angle for Physical Education = $\frac{10}{60}\times 360^{\circ }=60^{\circ }$

Angle for Maths = $\frac{12}{60}\times 360^{\circ }=72^{\circ }$

Angle for Science = $\frac{16}{60}\times 360^{\circ }=96^{\circ }$

Page No. 65Figure it Out

Q1: Which of these are in inverse proportion?

(i) 

Ans: x₁ = 40, x₂ = 80, x₃ = 25, x₄ = 16
y₁ = 20, y₂ = 10, y₃ = 32, y₄ = 50
x₁y₁ = 40 × 20 = 800
x₂y₂ = 80 × 10 = 800
x₃y₃ = 25 × 32 = 800
x₄y₄ = 16 × 50 = 800
So, x₁y₁ = x₂y₂ = x₃y₃ = x₄y₄ = 800
∴ x and y are in inverse proportion

(ii) 

Ans: x₁ = 40, x₂ = 80, x₃ = 25, x₄ = 16
y₁ = 20, y₂ = 10, y₃ = 12.5, y₄ = 8
x₁y₁ = 40 × 20 = 800
x₂y₂ = 80 × 10 = 800
x₃y₃ = 25 × 12.5 = 312.5
x₄y₄ = 16 × 8 = 128
So, x₁y₁ = x₂y₂ ≠ x₃y₃ ≠ x₄y₄
∴ x and y are not in inverse proportion.

(iii) 

Ans: x₁ = 30, x₂ = 90, x₃ = 150, x₄ = 10
y₁ = 15, y₂ = 5, y₃ = 3, y₄ = 45
x₁y₁ = 30 × 15 = 450
x₂y₂ = 90 × 5 = 450
x₃y₃ = 150 × 3 = 450
x₄y₄ = 10 × 45 = 450
So, x₁y₁ = x₂y₂ = x₃y₃ = x₄y₄ = 450
∴ x and y are in inverse proportion.

Q2: Fill in the empty cells if x and y are in inverse proportion.

Ans: 

$\therefore x\text{ and }y\text{ are in }\mathbf{\text{inverse}}\mathbf{\text{proportion.}}$

$\therefore 16\times 9=12\times y_2$

$\Rightarrow y_2=\frac{16\times 9}{12}=12$

$\therefore 16\times 9=x_3\times 48$

$\Rightarrow x_3=\frac{16\times 9}{48}=3$

And

$16\times 9=36\times y_4$

$\Rightarrow y_4=\frac{16\times 9}{36}=4\Rightarrow y_4=4$

Page No. 67Figure it Out

Q1: Which of the following pairs of quantities are in inverse proportion?

(i) The number of taps filling a water tank and the time taken to fill it.

Ans: More taps → Less time to fill the tank
Fewer taps → More time to fill the tank
The quantities change in opposite directions by the same factor.
If we double the no. of taps, the time taken becomes half.
Hence, they are in inverse proportion.

(ii) The number of painters hired and the days needed to paint a wall of fixed size.

Ans: More painters → Fewer days needed
Fewer painters → More days needed.
If we double the no. of painters, the work gets done in half the time.
Hence, they are in inverse proportion.

(iii) The distance a car can travel and the amount of petrol in the tank.

Ans: Petrol → It decreases
Distance → It increases
When distance increases, then petrol decreases, so they are in inverse proportion.

(iv) The speed of a cyclist and the time taken to cover a fixed route.

Ans: Higher speed → Less time taken
Lower speed → More time taken
For a fixed distance, if speed doubles, time becomes half.
Hence, they are in inverse proportion.

(v) The length of cloth bought and the price paid at a fixed rate per metre.

Ans: More cloth → More price to pay
Less cloth → Less price to pay
Both quantities decrease together and increase together, so they are in direct proportion.

(vi) The number of pages in a book and the time required to read it at a fixed reading speed.

Ans: More pages → More time to read
Fewer pages → Less time to read
Both quantities decrease together and increase together, so they are in direct proportion.

Q2: If 24 pencils cost ₹120, how much will 20 such pencils cost?

Ans: The number of pencils and the cost of pencils are in direct proportion. 

If x is the required cost, then

$\frac{24}{20}=\frac{\text{₹}120}{x}$

⇒ x × 24 = ₹ 120 × 20

⇒ x = ₹ 100

So, the cost of 20 such pencils is ₹ 100.

Q3: A tank on a building has enough water to supply 20 families living there for 6 days. If 10 more families move in there, how long will the water last? What assumptions do you need to make to work out this problem?

Ans: The number of families and the number of days are in inverse proportion.
Assumptions needed
(i) All families use the same amount of water.
(ii) Water usage per family per day is constant.
(iii) No additional water is added to the tank.
Let the water last for x days.
So, 20 × 6 = 30 × x
⇒ x = 4
So, the water will last for 4 days.

Q4: Fill in the average number of hours each living being sleeps in a day by looking at the charts. Select the appropriate hours from this list: 15, 2.5, 20, 8, 3.5, 13, 10.5, 18.

Ans: Common Sleep Patterns:
Average no. of hours a giraffe sleeps = 2.5 hours
Average no. of hours an elephant sleeps = 3.5 hours
Average no. of hours a boy sleeps = 8 hours
Average no. of hours a dog sleeps = 10.5 hours
Average no. of hours a cat sleeps = 13 hours
Average no. of hours a squirrel sleeps = 15 hours
Average no. of hours a snake sleeps = 18 hours
Average no. of hours a bat sleeps = 20 hours

Page No. 68

Q5: The pie chart shows the result of a survey carried out to find the modes of transport used by children to go to school. Study the pie chart and answer the following questions.

(i) What is the most common mode of transport?
(ii) What fraction of children travel by car?
(iii) If 18 children travel by car, how many children took part in the survey? How many children use taxis to travel to school?
(iv) By which two modes of transport are equal numbers of children travelling?

Ans: (i) The largest angle is 120°, which corresponds to the Bus. 

(ii) The fraction of children who travel by car

$\frac{30}{360}=\frac{1}{12}$

(iii) Let x be the total number of children who took part in the survey.

Then 18 = $\frac{1}{12}\times x$

⇒ x = 18 × 12 = 216

The number of children using taxis is 0, as taxis are not a category in this chart.

(iv) Cycle and two-wheeler (60°)

Q6: Three workers can paint a fence in 4 days. If one more worker joins the team, how many days will it take them to finish the work? What are the assumptions you need to make?

Ans: When the number of workers increases, the number of days needed to paint the fence decreases.
Assumptions Needed
(i) All workers work at the same speed/rate.
(ii) The work is uniformly distributed among all workers.
(iii) All workers work for the same number of hours each day.
So, the number of workers and the number of days are in inverse proportion.
Let x be the no. of days taken.
3 × 4 = 4 × x
⇒ x = 3
So, they will take 3 days to finish the work.

Q7: It takes 6 hours to fill 2 tanks of the same size with a pump. How long will it take to fill 5 such tanks with the same pump?

Ans: No. of hours and no. of tanks are in direct proportion. 

Let 5 such tanks take x hours.

62=x5

⇒ x = 15

So, 5 tanks will take 15 hours.

Q8: A given set of chairs are arranged in 25 rows, with 12 chairs in each row. If the chairs are rearranged with 20 chairs in each row, how many rows does this new arrangement have?

Ans: No. of rows and no. of chairs in each row are in inverse proportion.
Let the new arrangement have x rows.
25 × 12 = x × 20
⇒ x = 15
So, the new arrangement has 15 rows.

Q9: A school has 8 periods a day, each of 45 minutes duration. How long is each period, if the school has 9 periods a day, assuming that the number of school hours per day stays the same?

Ans: No. of periods and the duration of each period are in inverse proportion.
Let each period be of x minutes.
8 × 45 = 9 × x
⇒ x = 40
So, each period is 40 minutes

Q10: A small pump can fill a tank in 3 hours, while a large pump can fill the same tank in 2 hours. If both pumps are used together, how long will the tank take to fill?

Ans: Let x litres be the capacity of the tank.

Then, water filled by small tank in one hour = x3 litre

Water filled by large tank in one hour = x2 litre

So, total water filled by both pumps in one hour = (x3+x2) litre = 5×6 litre

∴ Time taken by both pumps used together to fill the tank = (1÷5×6)×x

= 1×65x×x

= 65 hour

= 115 hour

Q11: A factory requires 42 machines to produce a given number of toys in 63 days. How many machines are required to produce the same number of toys in 54 days?

Ans: No. of machines and no. of days are in inverse proportion.
42 × 63 = x × 54
⇒ x = 49
So, 49 machines are required.

Q12: A car takes 2 hours to reach a destination, travelling at a speed of 60 km/h. How long will the car take if it travels at a speed of 80 km/h?

Ans: Let the car take t hours.
The speed of the car and the time taken are in inverse proportion.
So, 2 × 60 = t × 80
⇒ t = 1.5 hour
So, the car will take 1.5 hours.

NCERT Solutions: The BAUDHĀYANA- Pythagoras Theorem

​Page No. 39

Figure it Out

Q1: Earlier, we saw a method to create a square with double the area of a given square paper. There is another method to do this in which two identical square papers are cut in the following way (pieces labeled 1, 2, 3, 4). 

Can you arrange these pieces to create a square with double the area of either square?

Ans: Given: Two identical squares
These two are cut diagonally, forming two equal triangles, as shown in the figure.

​Thus, we have four identical triangles from two identical squares.
Area of two triangles = Area of the given square
Area of four triangles = double the area of the given square.

Q2: The length of the two equal sides of an isosceles right triangle is given. Find the length of the hypotenuse. Find bounds on the length of the hypotenuse such that they have at least one digit after the decimal point.

(i) 3   (ii) 4   (iii) 6   (iv) 8   (v) 9

Ans: For an isosceles right triangle with equal sides of length a, the hypotenuse c is given by:

Formula: c² = 2a²  or  c = a$\sqrt{2}$

(i) When a = 3:

c = 3$\sqrt{2}$

To find bounds, we calculate:

• 4² = 16 and 5² = 25
• (3$\sqrt{2}$)² = $9 \times 2$ = 18
• Since 16 < 18 < 25, we have 4 < 3$\sqrt{2}$ < 5

For more precision:

• 4.2² = 17.64
• 4.3² = 18.49
• Since 17.64 < 18 < 18.49

Ans: Length of hypotenuse = 3$\sqrt{2}$ units ≈ 4.24 units
Bounds: 4.2 < hypotenuse < 4.3

(ii) When a = 4:

c = 4$\sqrt{2}$

To find bounds:

• (4$\sqrt{2}$)² = $16 \times 2$ = 32
• 5² = 25 and 6² = 36
• Since 25 < 32 < 36, we have 5 < 4$\sqrt{2}$ < 6

For more precision:

• 5.6² = 31.36
• 5.7² = 32.49
• Since 31.36 < 32 < 32.49

Ans: Length of hypotenuse = 4$\sqrt{2}$ units ≈ 5.66 units
Bounds: 5.6 < hypotenuse < 5.7

(iii) When a = 6:

c = 6$\sqrt{2}$

To find bounds:

• (6$\sqrt{2}$)² = $36 \times 2$ = 72
• 8² = 64 and 9² = 81
• Since 64 < 72 < 81, we have 8 < 6$\sqrt{2}$ < 9

For more precision:

• 8.4² = 70.56
• 8.5² = 72.25
• Since 70.56 < 72 < 72.25

Ans: Length of hypotenuse = 6$\sqrt{2}$ units ≈ 8.49 units
Bounds: 8.4 < hypotenuse < 8.5

(iv) When a = 8:

c = 8$\sqrt{2}$

To find bounds:

• (8$\sqrt{2}$)² = $64 \times 2$ = 128
• 11² = 121 and 12² = 144
• Since 121 < 128 < 144, we have 11 < 8$\sqrt{2}$ < 12

For more precision:

• 11.3² = 127.69
• 11.4² = 129.96
• Since 127.69 < 128 < 129.96

Ans: Length of hypotenuse = 8$\sqrt{2}$ units ≈ 11.31 units
Bounds: 11.3 < hypotenuse < 11.4

(v) When a = 9:

c = 9$\sqrt{2}$

To find bounds:

• (9$\sqrt{2}$)² = $81 \times 2$ = 162
• 12² = 144 and 13² = 169
• Since 144 < 162 < 169, we have 12 < 9$\sqrt{2}$ < 13

For more precision:

• 12.7² = 161.29
• 12.8² = 163.84
• Since 161.29 < 162 < 163.84

Ans: Length of hypotenuse = 9$\sqrt{2}$ units ≈ 12.73 units
Bounds: 12.7 < hypotenuse < 12.8

Page No. 40

Q3: The hypotenuse of an isosceles right triangle is 10. What are its other two sidelengths? [Hint: Find the area of the square composed of two such right triangles.]

Ans: 

Given: Hypotenuse of isosceles right triangle = 10 units

To find: Length of the two equal sides

Solution:

Let a be the length of each equal side.

Using the formula for isosceles right triangle: c² = 2a²

Given c = 10

Substituting: (10)² = 2a²

100 = 2a²

Dividing by 2: a² = $\frac{100}{2}$ = 50

Taking square root: a = $\sqrt{50}$

Simplifying $\sqrt{50}$:

$\sqrt{50}$ = $\sqrt{25 \times 2}$ = $\sqrt{25}$ × $\sqrt{2}$ = 5$\sqrt{2}$

Finding bounds:

• (5$\sqrt{2}$)² = $25 \times 2$ = 50
• 7² = 49 and 8² = 64
• Since 49 < 50 < 64, we have 7 < 5$\sqrt{2}$ < 8

More precisely:

• 7.0² = 49
• 7.1² = 50.41
• So 7.0 < 5$\sqrt{2}$ < 7.1

Ans: Each of the two equal sides has length 5$\sqrt{2}$ units ≈ 7.07 units

Page No. 47

Figure it Out

Q1: If a right-angled triangle has shorter sides of lengths 5 cm and 12 cm, then what is the length of its hypotenuse? First draw the right-angled triangle with these sidelengths and measure the hypotenuse, then check your answer using Baudhāyana’s Theorem.

Ans: Given AB = 5 cm
BC = 12 cm
AC = 13 cm (by measurement)

Using Baudhāyana’s Theorem

AC² = AB² + BC²

⇒ 5² + 12² = c²

⇒ 25 + 144 = c²

⇒ 169 = c²

⇒ c = $\sqrt{169}$ = 13

Answer: The length of the hypotenuse is 13 cm.

Q2: If a right-angled triangle has a short side of length 8 cm and hypotenuse of length 17 cm, what is the length of the third side? Again, try drawing the triangle and measuring, and then check your answer using Baudhāyana’s Theorem.

Ans: Here, AB = 8 cm
AC = 17 cm
BC = 15 cm (by measurement)

Using Baudhāyana’s Theorem,

AB² + BC² = AC²
8² + BC² = 17²
BC² = 289 – 64
= 225
= 15²
∴ BC = 15 cm
Therefore, the other side of the right-angled triangle is 15 cm, which satisfies Baudhayana’s Theorem.

Q3: Using the constructions you have now seen, how would you construct a square whose area is triple the area of a given square? Five times the area of a given square?

Ans: (a) ABCD is a square with side a.
AC = a√2
ACEF is a rectangle with sides a√2 and a.
Now AE = a√3

​AEGH is a square with a side of a√3
Then Ar AEGH = 3a²
Ar ABCD = a²
∴ Area of AEGH = 3 × Area of ABCD

​(b) ABCD and CFED are squares with side ‘a’.
BE is a diagonal of the rectangle ABFE.
In rectangle ABFE
EF = a and BF = BC + CF = a + a = 2a

​Using Baudhayana’s Theorem
BE² = EF² + BF²
= a² + (2a)²
= a² + 4a²
= 5a²
BE = √5 a
BEFG is a square with side BE.
Area BEFG = BE²
= (√5a)²
= 5a²
Then the area of BEFG = 5 × the area of ABCD

Q4: Let a, b and c denote the length of the sides of a right triangle, with c being the length of the hypotenuse. Find the missing sidelength in each of the following cases:

(i) a = 5, b = 7

Ans:

Given: a = 5, b = 7, c = ?

Formula: a² + b² = c²

Calculation:

• 5² + 7² = c²
• 25 + 49 = c²
• 74 = c²
• c = $\sqrt{74}$

Finding bounds:

• 8² = 64 and 9² = 81
• Since 64 < 74 < 81, we have 8 < $\sqrt{74}$ < 9

More precise:

• 8.6² = 73.96
• 8.7² = 75.69
• So 8.6 < $\sqrt{74}$ < 8.7

Ans: c = $\sqrt{74}$ ≈ 8.60 units

(ii) a = 8, b = 12

Ans:

Given: a = 8, b = 12, c = ?

Formula: a² + b² = c²

Calculation:

• 8² + 12² = c²
• 64 + 144 = c²
• 208 = c²
• c = $\sqrt{208}$

Simplifying:

• $\sqrt{208}$ = $\sqrt{16 \times 13}$ = 4$\sqrt{13}$

Finding bounds:

• 14² = 196 and 15² = 225
• Since 196 < 208 < 225, we have 14 < $\sqrt{208}$ < 15

More precise:

• 14.4² = 207.36
• 14.5² = 210.25
• So 14.4 < $\sqrt{208}$ < 14.5

Answer: c = $\sqrt{208}$ = 4$\sqrt{13}$ ≈ 14.42 units

(iii) a = 9, c = 15

Ans:

Given: a = 9, c = 15, b = ?

Formula: a² + b² = c²

Rearranging: b² = c² – a²

Calculation:

• b² = 15² – 9²
• b² = 225 – 81
• b² = 144
• b = $\sqrt{144}$ = 12

Answer: b = 12 units

Verification: 9² + 12² = 81 + 144 = 225 = 15²

(iv) a = 7, b = 12

Ans:

Given: a = 7, b = 12, c = ?

Formula: a² + b² = c²

Calculation:

• 7² + 12² = c²
• 49 + 144 = c²
• 193 = c²
• c = $\sqrt{193}$

Finding bounds:

• 13² = 169 and 14² = 196
• Since 169 < 193 < 196, we have 13 < $\sqrt{193}$ < 14

More precise:

• 13.8² = 190.44
• 13.9² = 193.21
• So 13.8 < $\sqrt{193}$ < 13.9

Answer: c = $\sqrt{193}$ ≈ 13.89 units

(v) a = 1.5, b = 3.5

Ans:

Given: a = 1.5, b = 3.5, c = ?

Formula: a² + b² = c²

Calculation:

• (1.5)² + (3.5)² = c²
• 2.25 + 12.25 = c²
• 14.5 = c²
• c = $\sqrt{14.5}$

Finding bounds:

• 3² = 9 and 4² = 16
• Since 9 < 14.5 < 16, we have 3 < $\sqrt{14.5}$ < 4

More precise:

• 3.8² = 14.44
• 3.9² = 15.21
• So 3.8 < $\sqrt{14.5}$ < 3.9

Answer: c = $\sqrt{14.5}$ ≈ 3.81 units

Page No. 48Math Talk

Q: List down all the Baudhāyana triples with numbers less than or equal to 20.

Ans: Baudhāyana triples with numbers ≤ 20:

To find these, we check which triples (a, b, c) satisfy a² + b² = c² where a, b, c ≤ 20.

The complete list:

1. (3, 4, 5)– Primitive
   • 3² + 4² = 9 + 16 = 25 = 5²
2. (6, 8, 10)– Scaled version of (3, 4, 5) [multiply by 2]
   • 6² + 8² = 36 + 64 = 100 = 10²
3. (5, 12, 13)– Primitive
   • 5² + 12² = 25 + 144 = 169 = 13²
4. (9, 12, 15)– Scaled version of (3, 4, 5) [multiply by 3]
   • 9² + 12² = 81 + 144 = 225 = 15²
5. (8, 15, 17)– Primitive
   • 8² + 15² = 64 + 225 = 289 = 17²
6. (12, 16, 20)– Scaled version of (3, 4, 5) [multiply by 4]
   • 12² + 16² = 144 + 256 = 400 = 20²

Total: 6 Baudhāyana triples with numbers ≤ 20

Primitive triples: (3, 4, 5), (5, 12, 13), (8, 15, 17)
Non-primitive triples: (6, 8, 10), (9, 12, 15), (12, 16, 20)

Q: Is there an unending sequence of Baudhāyana triples?

Ans: Yes, there is an unending (infinite) sequence of Baudhāyana triples.

Proof:

We know that (3, 4, 5) is a Baudhāyana triple.

We can generate infinite triples by multiplying each term by any positive integer k:

• ($3 \times 1$, $4 \times 1$, $5 \times 1$) = (3, 4, 5)
• ($3 \times 2$, $4 \times 2$, $5 \times 2$) = (6, 8, 10)
• ($3 \times 3$, $4 \times 3$, $5 \times 3$) = (9, 12, 15)
• ($3 \times 4$, $4 \times 4$, $5 \times 4$) = (12, 16, 20)
• And so on…

Since k can be any positive integer (1, 2, 3, 4, 5, …, ∞), and each gives us a valid Baudhāyana triple, there are infinitely many Baudhāyana triples.

Q: Is (30, 40, 50) a Baudhāyana triple?

Ans: Yes, (30, 40, 50) is a Baudhāyana triple.

Verification:

We need to check if 30² + 40² = 50²

Calculation:

• 30² = 900
• 40² = 1600
• 50² = 2500
• 900 + 1600 = 2500

Ans: Yes, (30, 40, 50) is a Baudhāyana triple.

Note: This is a scaled version of (3, 4, 5), multiplied by 10.

• ($3 \times 10$, $4 \times 10$, $5 \times 10$) = (30, 40, 50)

Q: Is (300, 400, 500) a Baudhāyana triple?

Ans: Yes, (300, 400, 500) is a Baudhāyana triple.

Verification:

We need to check if 300² + 400² = 500²

Calculation:

• 300² = 90,000
• 400² = 1,60,000
• 500² = 2,50,000
• 90,000 + 1,60,000 = 2,50,000

Answer: Yes, (300, 400, 500) is a Baudhāyana triple.

Note: This is a scaled version of (3, 4, 5), multiplied by 100.

• ($3 \times 100$, $4 \times 100$, $5 \times 100$) = (300, 400, 500)

Q: Do you see any pattern among the triples (3, 4, 5), (6, 8, 10), (9, 12, 15), (12, 16, 20)?

Ans: Yes, there is a clear pattern:

Observation:

• (3, 4, 5) = ($3 \times 1$, $4 \times 1$, $5 \times 1$)
• (6, 8, 10) = ($3 \times 2$, $4 \times 2$, $5 \times 2$)
• (9, 12, 15) = ($3 \times 3$, $4 \times 3$, $5 \times 3$)
• (12, 16, 20) = ($3 \times 4$, $4 \times 4$, $5 \times 4$)

Pattern: Each triple is obtained by multiplying all three numbers of (3, 4, 5) by the same positive integer.

General form: (3k, 4k, 5k) where k = 1, 2, 3, 4, …

This pattern shows that:

1. All these triples are scaled versions of the primitive triple (3, 4, 5)
2. If we know one Baudhāyana triple, we can generate infinitely many more by scaling
3. Among these, only (3, 4, 5) is primitive (no common factor > 1)

Q: Can we form a conjecture on Baudhāyana triples based on this observation?

Ans: Yes, we can form the following conjecture:

Conjecture: (3k, 4k, 5k) is a Baudhāyana triple, where k is any positive integer.

Verification of the conjecture:

We need to check if (3k)² + (4k)² = (5k)²

Left side:

• (3k)² + (4k)²
• = 9k² + 16k²
• = 25k²

Right side:

• (5k)² = 25k²

Since LHS = RHS, the equation is satisfied!

Conclusion: The conjecture is TRUE.

(3k, 4k, 5k) is indeed a Baudhāyana triple for any positive integer k.

This proves there are infinitely many Baudhāyana triples (one for each value of k = 1, 2, 3, …).

Q: Can we further generalise the conjecture?

Ans: Yes, we can make a more general statement:

Page No. 49

Q: Is (5, 12, 13) a primitive Baudhāyana triple? What are the other primitive Baudhāyana triples with numbers less than or equal to 20?

Ans: Yes, (5, 12, 13) is a primitive Baudhāyana triple.

Reason:

• First, verify it’s a Baudhāyana triple: 5² + 12² = 25 + 144 = 169 = 13²
• Check for common factors: The numbers 5, 12, and 13 have no common factor greater than 1.
• GCD(5, 12, 13) = 1

Therefore, (5, 12, 13) is primitive.

From our earlier list, the primitive triples are:

1. (3, 4, 5)

• GCD(3, 4, 5) = 1

2. (5, 12, 13)

• GCD(5, 12, 13) = 1

3. (8, 15, 17)

• GCD(8, 15, 17) = 1

There are 3 primitive Baudhāyana triples with numbers ≤ 20:

• (3, 4, 5)
• (5, 12, 13)
• (8, 15, 17)

Q: Generate 5 scaled versions of each of these primitive triples. Are these scaled versions primitive?

Ans: Scaled versions of (3, 4, 5):

1. k = 2: (6, 8, 10)
2. k = 3: (9, 12, 15)
3. k = 4: (12, 16, 20)
4. k = 5: (15, 20, 25)
5. k = 6: (18, 24, 30)

Are these primitive? No, each has common factor k > 1.

Scaled versions of (5, 12, 13):

1. k = 2: (10, 24, 26)
2. k = 3: (15, 36, 39)
3. k = 4: (20, 48, 52)
4. k = 5: (25, 60, 65)
5. k = 6: (30, 72, 78)

Are these primitive? No, each has common factor k > 1.

Scaled versions of (8, 15, 17):

1. k = 2: (16, 30, 34)
2. k = 3: (24, 45, 51)
3. k = 4: (32, 60, 68)
4. k = 5: (40, 75, 85)
5. k = 6: (48, 90, 102)

Are these primitive? No, each has common factor k > 1.

Conclusion: No, scaled versions are never primitive because they all have a common factor k > 1.

Q: If (a, b, c) is non-primitive, and the integers have f — greater than 1 — as a common factor, then is (a/f, b/f, c/f) a Baudhāyana triple? Check this statement for (9, 12, 15). Justify this statement.

Ans: Yes, if (a, b, c) is non-primitive with common factor f > 1, then (a/f, b/f, c/f) is also a Baudhāyana triple.

Checking for (9, 12, 15):

First, find the common factor:

• 9 = $3 \times 3$
• 12 = $3 \times 4$
• 15 = $3 \times 5$
• Common factor f = 3

Dividing by f = 3:

• ($\frac{9}{3}$, $\frac{12}{3}$, $\frac{15}{3}$) = (3, 4, 5)

Verification:

• 3² + 4² = 9 + 16 = 25 = 5²

Yes, (3, 4, 5) is a Baudhāyana triple!

General Justification:

Given: (a, b, c) is a Baudhāyana triple with common factor f

• This means: a² + b² = c²
• Also: a = f × p, b = f × q, c = f × r for some integers p, q, r

To prove: (a/f, b/f, c/f) = (p, q, r) is a Baudhāyana triple

Starting with: a² + b² = c²

Substituting:

• (f × p)² + (f × q)² = (f × r)²
• f²p² + f²q² = f²r²
• f²(p² + q²) = f²r²

Dividing both sides by f²:

• p² + q² = r²

This proves: (p, q, r) = (a/f, b/f, c/f) is a Baudhāyana triple!

Important conclusion:

• Every non-primitive triple can be “reduced” to a primitive triple by dividing by the common factor
• If we find all primitive triples, we can generate all Baudhāyana triples by scaling

Page No. 50

Figure it Out

Q1: Find 5 more Baudhāyana triples using this idea.

Ans: (1 + 3 + 5 + … + 47) + 49 = 25²
24² + 7² = 25² (24, 7, 25)
(1 + 3 + 5 + … + 79) + 81 = 41²
40² + 9² = 41² (40, 9, 41)
(1 + 3 + 5 + … + 119) + 121 = 61²
60² + 11² = 61² (60, 11, 61)
(1 + 3 + 5 + … + 167) + 169 = 85²
842 + 132 = 852 (84, 13, 85)
(1 + 3 + 5 + …+ 223) + 225 = 113²
112² + 15² = 113² (112, 15, 113)

Q2: Does this method yield non-primitive Baudhāyana triples? [Hint: Observe that among the triples generated, one of the smaller sidelengths is one less than the hypotenuse.]

Ans: (24, 7, 25)
HCF of 24, 7, 25 is 1.
(40, 9, 41)
HCF of 40, 9, 41 is 1, etc.
The triples generated by the above method are primitive in nature.

Q3: Are there primitive triples that cannot be obtained through this method? If yes, give examples.

Ans: In each of the above case we have taken the sum of the first ‘n’ odd numbers where ‘n’ is a perfect square.
We observe that the smallest number in the triple is always odd.
Consider triples such as (8, 15, 17), (16, 63, 65), etc.
Such triples cannot be generated by this method.

Figure it Out

Q1: Find the diagonal of a square with sidelength 5 cm.

Ans:

Given: Square with side = 5 cm

To find: Length of diagonal

Method:

A diagonal of a square divides it into two congruent right-angled triangles.

For each triangle:

• Both perpendicular sides = 5 cm (sides of square)
• Hypotenuse = diagonal of square

Using Baudhāyana’s Theorem:

Let d = length of diagonal

• 5² + 5² = d²
• 25 + 25 = d²
• 50 = d²
• d = $\sqrt{50}$

Simplifying:

• $\sqrt{50}$ = $\sqrt{25 \times 2}$ = $\sqrt{25}$ × $\sqrt{2}$ = 5$\sqrt{2}$

Finding approximate value:

• $\sqrt{2}$ ≈ 1.414
• 5$\sqrt{2}$ ≈ $5 \times 1$.414 = 7.07

Ans: The diagonal of the square is 5$\sqrt{2}$ cm ≈ 7.07 cm.

Q2: Find the missing sidelengths in the following right triangles:

(i) Sides: 7 and ?, Hypotenuse: ?

If one perpendicular side = 7 and other perpendicular side = 9:

Given: a = 7, b = 9, c = ?

Using Baudhāyana’s Theorem:

• 7² + 9² = c²
• 49 + 81 = c²
• 130 = c²
• c = $\sqrt{130}$

(ii) Sides: 4 and 10, Hypotenuse: ?

Given: a = 4, b = 10, c = ?

Using Baudhāyana’s Theorem:

• 4² + 10² = c²
• 16 + 100 = c²
• 116 = c²
• c = $\sqrt{116}$

Simplifying:

• $\sqrt{116}$ = $\sqrt{4 \times 29}$ = 2$\sqrt{29}$

(iii) Sides: 40 and ?, Hypotenuse: 41

Given: a = 40, c = 41, b = ?

Using Baudhāyana’s Theorem:

b² = c² – a²

• b² = 41² – 40²
• b² = 1681 – 1600
• b² = 81
• b = $\sqrt{81}$ = 9

Ans: b = 9 units

(iv) Sides: 27 and ?, Hypotenuse: 45

Given: a = 27, c = 45, b = ?

Using Baudhāyana’s Theorem:

b² = c² – a²

• b² = 45² – 27²
• b² = 2025 – 729
• b² = 1296
• b = $\sqrt{1296}$ = 36

Ans: b = 36 units

(v) Sides: √200 and 10, Hypotenuse: ?

10² + d² = (√200)²
⇒ 100 + d² = 200
⇒ d² = 200 – 100
⇒ d² = 100
⇒ d = √100
⇒ d = 10

(vi) Sides: 10 and 150, Hypotenuse: ?

​e² = 10² + (√150)²
⇒ e² = 100 + 150
⇒ e² = 250
⇒ e = √250
⇒ e =  5×5×5×2−−−−−−−−−−√
⇒ e = 5√10

Q3: Find the sidelength of a rhombus whose diagonals are of length 24 units and 70 units.

Ans:

Given:

• Diagonal 1 (d₁) = 24 units
• Diagonal 2 (d₂) = 70 units

To find: Side length of the rhombus

Key properties of a rhombus:

1. Diagonals bisect each other at right angles (90°)
2. All four sides are equal

Solution:

When diagonals intersect, they form 4 right-angled triangles.

Each right triangle has:

• One side = d₁/2 = $\frac{24}{2}$ = 12 units
• Other side = d₂/2 = $\frac{70}{2}$ = 35 units
• Hypotenuse = side of rhombus (s)

Using Baudhāyana’s Theorem:

s² = 12² + 35²

• s² = 144 + 1225
• s² = 1369
• s = $\sqrt{1369}$ = 37

Ans: The side length of the rhombus is 37 units.

Q4: Is the hypotenuse the longest side of a right triangle? Justify your answer.

Ans: c² = a² + b²
∴ c² > a² and c² > b²
or c > a and c > b
Hence, ‘c’ is the longest side of the right triangle.

Q5: True or False—Every Baudhāyana triple is either a primitive triple or a scaled version of a primitive triple.

Ans:

TRUE

Explanation:

Every Baudhāyana triple falls into one of two categories:

Category 1: Primitive Triples

• These have no common factor greater than 1
• Examples: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25)
• GCD of all three numbers = 1

Category 2: Non-primitive Triples (Scaled versions)

• These have a common factor f > 1
• They can be reduced by dividing by f
• The reduced form is a primitive triple
• Examples:
  • (6, 8, 10) = 2 × (3, 4, 5) → scaled version of (3, 4, 5)
  • (9, 12, 15) = 3 × (3, 4, 5) → scaled version of (3, 4, 5)
  • (10, 24, 26) = 2 × (5, 12, 13) → scaled version of (5, 12, 13)

Proof:

For any Baudhāyana triple (a, b, c):

• Let f = GCD(a, b, c) (the greatest common divisor)

Case 1: If f = 1

• The triple is primitive

Case 2: If f > 1

• We can write: a = f × p, b = f × q, c = f × r
• Then (p, q, r) = (a/f, b/f, c/f) is a primitive triple
• And (a, b, c) = f × (p, q, r) is a scaled version

Therefore, the statement is TRUE.

Every Baudhāyana triple is either primitive OR a scaled version of a primitive triple. There’s no third category!

Q6: Give 5 examples of rectangles whose sidelengths and diagonals are all integers.

Ans: 

Q7: Construct a square whose area is equal to the difference of the areas of squares of sidelengths 5 units and 7 units.

Ans: Area of square = 7² – 5²
= 49 – 25
= 24 sq. units

​1. Construct a square ABCD with a side of 2 cm.
Then DB = 2√2 units
2. Draw BE ⊥ DB at B such that BE = 4 units
3. Join DE.
DE² = (2√2)² + 4²
= 8 + 16
= 24
⇒ DE = √24
4. Draw a square with side DE (DEFG).

​Area of DEFG = (√24)² = 24 square units.

Q8: (i) Using the dots of a grid as the vertices, can you create a square that has an area of (a) 2 sq. units, (b) 3 sq. units, (c) 4 sq. units, and (d) 5 sq. units?

​(ii) Suppose the grid extends indefinitely. What are the possible integer-valued areas of squares you can create in this manner?

Ans: (i) (a) Area = 2 sq. units

​2 = 1² + 1²

Mark dots A, B, C, and D as shown.
Join AB, BC, CD, and DA.
Then ABCD is a square and area ABCD = 2 sq. units
(b) Square with area 3 units is not possible as 3 ≠ a² + a² for any integer ‘a’.
(c) 4 = 2 × 2

Mark dots A, B, C, D as shown.
Join AB, BC, CD, DA.
Then ABCD is a square.
and ar ABCD = 2 × 2 = 4 sq. units
(d) (i) 

Mark dots A, B, C, and D as shown.
Join A, B, C, and D
AB² = 2² + 1² = 5
AB = √5 units
Hence, ABCD is a square with an area of 5 sq units.

(ii) Let the given value of area be x, where ‘x’ is an integer.
Then, x = a² + b², where ‘a’ and ‘b’ are integers or x is a perfect square, we can create squares with vertices as dots of the grid.

Page No. 54

Q9: Find the area of an equilateral triangle with sidelength 6 units. [Hint: Show that an altitude bisects the opposite side. Use this to find the height.]

Ans: Let ΔABC be an equilateral triangle.
AB = BC = CA = 6 cm

Let AD be perpendicular to BC.
Then ∠1 = ∠2 (each = 90°)
AB = AC (each = 6 cm)
AD = AD (common)
ΔADB ≅ ΔADC (RHS)
BD = DC (CPCT)
∴ BD = DC = 1/212 × 6 cm = 3 cm
In ΔADC,
h² + 3² = 6² (Baudhayana’s triple)
⇒ h² = 36 – 9 = 27
⇒ h = 3×3×3−−−−−−−√
⇒ h = 3√3 cm
Ar ABC = 1/21​× BC × AD
=  1/21​× 6 × 3√3 sq. units
= 9√3 sq. units