Text Book Solutions All Class

Q1. Give three examples of data which you can get from your day-to-day life.
Sol:
Here are the three examples which are related to our day-to-day life:

• The number of boys in a sports team.
• Electricity bills for last one year.
• The number of students appearing for board exams at your school.

Q2: The number of family members in 10 flats of society are
2, 4, 3, 3, 1,0,2,4,1,5.
Find the mean number of family members per flat.
Sol:
Number of family members in 10 flats -2, 4, 3, 3, 1, 0, 2, 4, 1, 5.
So, we get,
Mean = sum of observation/ total no of observations
Mean = (2 + 4+ 3 + 3 + 1 + 0 + 2 + 4 + 1 + 5) / 10
Mean = 25/10 = 2.5

Q3: The daily minimum questions solved by a student during a week were as under:

Find the mean.
Sol:
Number of questions solved in a week: 35, 30, 27, 32, 23, 28.
So, we get,
Mean = sum of observation/ total no of observations = (35+30+27+32+23+28) / 6 = 175/6 = 29.167

Q4: The mean weight of a class of 34 students is 46.5 kg. If the weight of the new boy is included, the mean is rises by 500 g. Find the weight of the new boy.
Sol:
The mean weight of 34 students = 46.5
Sum of the weight of 34 students = (46.5 * 34) = 1581
Change or increase in the mean weight when the weight of a new boy is added = 0.5
So, the new mean = (46.5 +0.5) = 47
So, let the weight of the new boy be y.
So, (sum of weight of 34 students + weight of new boy) / 35 = 47
(1581+ y)/ 35 = 47
1581 + y = 1645
y = 1645 – 1581 = 64

Q5: The height of 20 students of class V are noted as follows

4, 4.5, 5, 5.5, 4, 4, 4.5, 5, 5.5, 4, 3.5, 3.5, 4.2, 4.6, 4.2, 4.7, 5.5, 5.3, 5, 5.5.

• Make a frequency distribution table for the above data.
• Which is the most common height and which is the rarest height among these students?

Sol:

1. The required frequency distribution table is:

2. The most common heights are 4 and 5.5.
The rarest heights are 4.6 and 4.7.

Q6: The following is the list of number of coupons issued in a school canteen during a week:
105, 216, 322, 167, 273, 405 and 346.
Find the average no. of coupons issued per day.
Sol:
Number of coupons issued in a week: 105, 216, 322, 167, 273, 405 and 346.
So, we get,
Mean = sum of observation/ total no of observations
Mean = (106+ 215+ 323+166+273+405+346)/ 7 = 1834/7
Mean = 262

Q7: If the mean of six observations y, y + 1, y + 4, y + 6, y + 8, y + 5 is 13, find the value of y.
Sol:
Mean = sum of observation/ total no of observations
13 = (y + y + 1+ y + 4+ y + 6+ y + 8+ y + 5) / 6
13 = (6y + 24)/6
(13 * 6) = 6y +24
(13 * 6) – 24 = 6y
(13 * 6) – 6 * 4 = 6y
6(13 – 4) = 6y
y = 9

Q8: The Number of books issued to 13 students in a class are:
25, 19, 24, 23, 29, 31, 19, 20, 22, 26, 17, 35, 21.
Find the median no. of books for the above data.
Sol:
Let’s arrange the data given in ascending order – 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29,31,35.
n= 13, so it’s an odd number
Median = (n+1) / 2 observations
= (13+1)/ 2 = (14/2)th observation = 7th observation = 23

Q9: The weight (in kg) of 7 students of a class are 44, 52, 55, 60, 50, 49, 45.
Find the median weight.
Sol:
Let’s arrange the data given in ascending order – 44, 45, 49, 50, 52, 55, 60.
n= 7, so it’s an odd number
Median = (n+1) / 2 observations
= (7+1)/ 2 = (8/2)th observation = 4th observation = 50 kg

Q1: In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
Sol:
Given,
Length of the cylindrical pipe = h = 28 m
Diameter of the pipe = 5 cm
Now, the radius of piper (r) = 5/ 2 cm = 2.5 cm = 0.025 m
Total radiating surface in the system = Total surface area of the cylinder  
= 2πr(h + r)  
= 2 × (22/7) × 0.025 (28 + 0.025) m² 
= (44 x 0.025 x 28.025)/7 m²
= 4.4 m² (approx)

Q2: Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Sol:
Given,
Diameter of the cone = 24 m
Radius of the cone (r) = 24/2 = 12 m
Slant height of the cone (l) = 21 m
Total surface area of a cone = πr(l + r)
= (22/7) × 12 × (21 + 12)
= (22/7) × 12 × 33
= 1244.57 m²

Q3: The hollow sphere, in which the circus motorcyclist performs his stunts, has a diameter of 7 m. Find the area available to the motorcyclist for riding.
Sol:
Given,
Diameter of the sphere = 7 m
Radius (r) = 7/2 = 3.5 m
Now, the riding space available for the motorcyclist = Surface area of the sphere
= 4πr²
= 4 × (22/7) × 3.5 × 3.5
= 154 m²  

Q4: Hameed has built a cubical water tank with a lid for his house, with each outer edge 1.5 m long. He gets the outer surface of the tank excluding the base, covered with square tiles of side 25 cm (see in the figure below). Find how much he would spend on the tiles if the cost of the tiles is Rs.360 per dozen.

Sol:
Given,
Edge of the cubical tank (a) = 1.5 m = 150 cm
So, surface area of the tank = 5 × 150 × 150 cm²
The measure of side of a square tile = 25 cm
Area of each square tile = side × side = 25 × 25 cm²
Required number of tiles = (Surface area of the tank)/(area of each tile)
= (5 × 150 × 150)/(25 × 25)
= 180
Also, given that the cost of the tiles is Rs. 360 per dozen.
Thus, the cost of each tile = Rs. 360/12 = Rs. 30
Hence, the total cost of 180 tiles = 180 × Rs. 30 = Rs. 5400

Q5: The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of whitewashing the walls of the room and the ceiling at the rate of  Rs.7.50 per sq.m.
Sol:
Given,
Length of the room (l) = 5 m
Breadth of the room (b) = 4 m
Height of the room (h) = 3 m
Area of walls of the room = Lateral surface area of cuboid
= 2h(l + b)
= 2 × 3(5 + 4)
= 6 × 9
= 54 sq.m
Area of ceiling = Area of base of the cuboid = lb
= 5 × 4
= 20 sq.m
Area to be white washed = (54 + 20) sq.m = 74 sq.m
Given that, the cost of white washing 1 sq.m = Rs. 7.50
Therefore, the total cost of white washing the walls and ceiling of the room = 74 × Rs. 7.50 = Rs. 555

Q6: Curved surface area of a right circular cylinder is 4.4 sq.m. If the radius of the base of the cylinder is 0.7 m, find its height.
Sol:
Let h be the height of the cylinder.
Given,
Radius of the base of the cylinder (r) = 0.7 m
Curved surface area of cylinder = 4.4 m²
Thus,
2πrh = 4.4  
2 × 3.14 × 0.7 × h = 4.4
4.4 × h = 4.4
h = 4.4/4.4
h = 1  
Therefore, the height of the cylinder is 1 m.

Q7: The height of a cone is 16 cm and its base radius is 12 cm. Find the curved surface area and the total surface area of the cone. (Take π = 3.14)
Sol:
Given
Height of a cone (h) = 16 cm
Radius of the base (r) = 12 cm
Now,
Slant height of cone (l) = √(r² + h²)
= √(256 + 144)
= √400
= 20 cm
Curved surface area of cone = πrl
= 3.14 × 12 × 20 cm²
= 753.6 cm²
Total surface area = πrl + πr²
= (753.6 + 3.14 × 12 × 12) cm²
= (753.6 + 452.16) cm²
= 1205.76 cm²

Q8: The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs.210 per 100 sq.m.
Sol:
Given,
Slant height of a cone (l) = 25 m  
Diameter of the base of cone = 2r = 14 m  
∴ Radius = r = 7 m  
Curved Surface Area = πrl
= (22/7) x 7 x 25
= 22 × 25  
= 550 sq.m  
Also, given that the cost of white-washing 100 sq.m = Rs. 210
Hence, the total cost of white-washing for 550 sq.m = (Rs. 210 × 550)/100 = Rs. 1155

Q9: The paint in a certain container is sufficient to paint an area equal to 9.375 sq.m. How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?
Sol:
Given,
Dimensions of the brick = 22.5 cm × 10 cm × 7.5 cm
Here, l = 22.5 cm, b = 10 cm, h = 7.5 cm
Surface area of 1 brick  = 2(lb + bh + hl)
= 2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5) cm²
= 2(225 + 75 + 168.75) cm² 
= 2 x 468.75 cm² 
= 937.5 cm² 
Area that can be painted by the container = 9.375 m² (given)
= 9.375 × 10000 cm²
= 93750 cm²
Thus, the required number of bricks = (Area that can be painted by the container)/(Surface area of 1 brick)
= 93750/937.5
= 937500/9375
= 100

Q10: The curved surface area of a right circular cylinder of height 14 cm is 88 sq.cm. Find the diameter of the base of the cylinder.
Sol:
Let d be the diameter and r be the radius of a right circular cylinder.
Given,
Height of cylinder (h) = 14 cm
Curved surface area of right circular cylinder = 88 cm2
⇒ 2πrh = 88 cm²
⇒ πdh = 88 cm² (since d = 2r)
⇒ 22/7 x d x 14 cm = 88 cm²
⇒ d = 2 cm
Hence, the diameter of the base of the cylinder is 2 cm.

Q1: Find the area of a triangle whose two sides are 18 cm and 10 cm and the perimeter is 42cm.
Sol: Assume that the third side of the triangle to be “x”.
Now, the three sides of the triangle are 18 cm, 10 cm, and “x” cm
It is given that the perimeter of the triangle = 42cm
So, x = 42 – (18 + 10) cm = 14 cm
∴ The semi perimeter of triangle = 42/2 = 21 cm
Using Heron’s formula,

Area of the triangle,
= √[s (s-a) (s-b) (s-c)]
= √[21(21 – 18) (21 – 10) (21 – 14)] cm²
= √[21 × 3 × 11 × 7] m²
= 21√11 cm²

Q2: A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.
Sol: First, draw a line segment BE parallel to the line AD. Then, from B, draw a perpendicular on the line segment CD.
Now, it can be seen that the quadrilateral ABED is a parallelogram. So,AB = ED = 10 m
AD = BE = 13 m
EC = 25 – ED = 25 – 10 = 15 m
Now, consider the triangle BEC,
Its semi perimeter (s) = (13+ 14 + 15)/2 = 21 m
By using Heron’s formula,
Area of ΔBEC =
= √[s(s − a)(s − b)(s − c)]
= √(21 × (21 − 13) × (21 − 14) × (21 − 15)) m²
= √(21 × 8 × 7 × 6) m²
= 84 m²
We also know that the area of ΔBEC = (½) × CE × BF
84 cm² = (½) × 15 × BF
⇒ BF = (168/15) cm = 11.2 cm
So, the total area of ABED will be BF × DE, i.e. 11.2 × 10 = 112 m²
∴ Area of the field = 84 + 112 = 196 m²

Q3: Find the cost of laying grass in a triangular field of sides 50 m, 65 m and 65 m at the rate of Rs 7 per m².
Sol: According to the question,
Sides of the triangular field are 50 m, 65 m and 65 m.
Cost of laying grass in a triangular field = Rs 7 per m²
Let a = 50, b = 65, c = 65
s = (a + b + c)/2
⇒ s = (50 + 65 + 65)/2
= 180/2
= 90.
Area of triangle = √(s(s-a)(s-b)(s-c))
= √(90(90 – 50)(90 – 65)(90 – 65))
= √(90 × 40 × 25 × 25)
= 1500m²
Cost of laying grass = Area of triangle × Cost per m²
= 1500 × 7
= Rs.10500

Q4: How much paper of each shade is needed to make a kite given in the figure, in which ABCD is a square with diagonal 44 cm.

Sol: According to the figure,
AC = BD = 44cm
AO = 44/2 = 22cm
BO = 44/2 = 22cm
From ΔAOB,
AB2 = AO² + BO²
⇒ AB² = 22² + 22²
⇒ AB² = 2 × 22²
⇒ AB = 22√2 cm
Area of square = (Side)²
= (22√2)²
= 968 cm²
Area of each triangle (I, II, III, IV) = Area of square /4
= 968 /4
= 242 cm²
To find area of lower triangle,
Let a = 20, b = 20, c = 14
s = (a + b + c)/2
⇒ s = (20 + 20 + 14)/2 = 54/2 = 27.
Area of the triangle = √[s(s-a)(s-b)(s-c)]
= √[27(27 – 20)(27 – 20)(27 – 14)]
= √[27 × 7 × 7 × 13]
= 131.14 cm²
Therefore,
We get,
Area of Red = Area of IV
= 242 cm²
Area of Yellow = Area of I + Area of II
= 242 + 242
= 484 cm²
Area of Green = Area of III + Area of the lower triangle
= 242 + 131.14
= 373.14 cm²

Q5: The sides of a triangle are in the ratio of 12: 17: 25 and its perimeter is 540cm. Find its area.
Sol: The ratio of the sides of the triangle is given as 12: 17: 25
Now, let the common ratio between the sides of the triangle be “x”
∴ The sides are 12x, 17x and 25x
It is also given that the perimeter of the triangle = 540 cm
12x + 17x + 25x = 540 cm
⇒ 54x = 540cm
So, x = 10
Now, the sides of the triangle are 120 cm, 170 cm, 250 cm.
So, the semi perimeter of the triangle (s) = 540/2 = 270 cm
Using Heron’s formula,
Area of the triangle
= √[s(s − a)(s − b)(s − c)]
= √(270(270 − 120)(270 − 170)(270 − 250)) cm²
= √(270 × 150 × 100 × 20) cm²
= 9000 cm²

Q6: A rhombus-shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
Sol: Draw a rhombus-shaped field first with the vertices as ABCD. The diagonal AC divides the rhombus into two congruent triangles which are having equal areas. The diagram is as follows.

Consider the triangle BCD,
Its semi-perimeter = (48 + 30 + 30)/2 m = 54 m
Using Heron’s formula,
Area of the ΔBCD =
= √[s(s − a)(s − b)(s − c)]
= √(54(54 − 48)(54 − 30)(54 − 30)) m²
= √(54 × 6 × 24 × 24) m²
= 432 m²
∴ Area of field = 2 × area of the ΔBCD = (2 × 432) m² = 864 m²
Thus, the area of the grass field that each cow will be getting = (864/18) m² = 48 m²

Q7: The perimeter of an isosceles triangle is 32 cm. The ratio of the equal side to its base is 3: 2. Find the area of the triangle.
Sol: According to the question,
The perimeter of the isosceles triangle = 32 cm
It is also given that,
Ratio of equal side to base = 3 : 2
Let the equal side = 3x
So, base = 2x
Perimeter of the triangle = 32
⇒ 3x + 3x + 2x = 32
⇒ 8x = 32
⇒ x = 4.
Equal side = 3x = 3×4 = 12
Base = 2x = 2×4 = 8
The sides of the triangle = 12cm, 12cm and 8cm.
Let a = 12, b = 12, c = 8
s = (a + b + c)/2
⇒ s = (12 + 12 + 8)/2
= 32/2
= 16.
Area of the triangle = √(s(s – a)(s – b)(s – c))
= √(16(16 – 12)(16 – 12)(16 – 8))
= √(16 × 4 × 4 × 8)
= 32√2 cm²

Q8: A rectangular plot is given for constructing a house, having a measurement of 40 m long and 15 m in the front. According to the laws, a minimum of 3 m, wide space should be left in the front and back each and 2 m wide space on each of other sides. Find the largest area where a house can be constructed.
Sol:

Let the given rectangle be rectangle PQRS,
According to the question,
PQ = 40m and QR = 15m
As 3m is left in both front and back,
AB = PQ -3 -3
⇒ AB = 40 -6
⇒ AB = 34m
Also,
Given that 2m has to be left at both the sides,
BC = QR -2 – 2
⇒ BC = 15 -4
⇒ BC = 11m
Now, Area left for house construction is the area of ABCD.
Hence,
Area(ABCD) = AB × CD
= 34 × 11
= 374 m²
 

Q9:  Find the area of a triangle whose sides are respectively 150 cm, 120 cm, and 200 cm?

Sol: The triangle whose sides are:
a = 150 cm, b = 120 cm, c = 200 cm
The area of a triangle = √[s(s − a)(s − b)(s − c)]
Here, s = semi-perimeter of the triangle
2s = a + b + c
s = a + b + c2 = 150 + 200 + 1202 = 235 cm
Area of triangle = √[s(s − a)(s − b)(s − c)]
= √(235(235 − 150)(235 − 200)(235 − 120)) cm²
= √(235 × 85 × 35 × 115) cm²
= 8966.56 cm²

Q10: A triangular park has a perimeter of 300 m, and all its sides are equal in length. Find the area of the park using Heron’s formula.
Sol: Let each side of the equilateral triangle be $a$a.
The semi-perimeter of the triangle,

s = a + a + a2 = 3a2
Using Heron’s formula:
Area = √[s(s − a)(s − b)(s − c)] = √[s(s − a)³]
Area = √[ 3a2 × 3a2 − a ]³
Area = √34 a²
Now, the perimeter = 300 m
a + a + a = 300 ⇒ 3a = 300 ⇒ a = 100 m
Thus, area of park = √34 (100)² = 2500√3 m²

Q1: Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90° – (½)A, 90° – (½)B and 90° – (½)C.
Sol: Consider the following diagram:
Here, ABC is inscribed in a circle with center O and the bisectors of ∠A, ∠B and ∠C intersect the circumcircle at D, E and F respectively.
Now, join DE, EF and FD
As angles in the same segment are equal, so,
∠FDA = ∠FCA ————-(i)
∠FDA = ∠EBA ————-(i)
Adding equations (i) and (ii) we have,
∠FDA + ∠EDA = ∠FCA + ∠EBA
Or, ∠FDE = ∠FCA + ∠EBA = (½)∠C + (½)∠B
We know, ∠A + ∠B + ∠C = 180°
So, ∠FDE = (½)[∠C + ∠B] = (½)[180° – ∠A]
⇒ ∠FDE = [90 – (∠A/2)]
In a similar way,
∠FED = [90 – (∠B/2)]
And,
∠EFD = [90 – (∠C/2)]

Q2: Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Sol:
To prove: A circle drawn with Q as centre, will pass through A, B and O (i.e. QA = QB = QO)
Since all sides of a rhombus are equal,
AB = DC
Now, multiply (½) on both sides
(½)AB = (½)DC
So, AQ = DP
⇒ BQ = DP
Since Q is the midpoint of AB,
AQ= BQ
Similarly,
RA = SB
Again, as PQ is drawn parallel to AD,
RA = QO
Now, as AQ = BQ and RA = QO we have,
QA = QB = QO (hence proved).

Q3: If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lies on the third side.
Sol: First, draw a triangle ABC and then two circles having a diameter as AB and AC respectively.
We will have to now prove that D lies on BC and BDC is a straight line.

Proof:
As we know, angle in the semi-circle are equal
So, ∠ADB = ∠ADC = 90°
Hence, ∠ADB + ∠ADC = 180°
∴ ∠BDC is a straight line.
So, it can be said that D lies on the line BC.

Q4: ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
Sol: Consider the following diagram.
Consider the chord CD,
As we know, angles in the same segment are equal.
So, ∠CBD = ∠CAD
∴ ∠CAD = 70°
Now, ∠BAD will be equal to the sum of angles BAC and CAD.
So, ∠BAD = ∠BAC + ∠CAD
= 30° + 70°
∴ ∠BAD = 100°
As we know, the opposite angles of a cyclic quadrilateral sum up to 180 degrees.
So,
∠BCD + ∠BAD = 180°
Since, ∠BAD = 100°
So, ∠BCD = 80°
Now consider the ΔABC.
Here, it is given that AB = BC
Also, ∠BCA = ∠CAB (Angles opposite to equal sides of a triangle)
∠BCA = 30°
also, ∠BCD = 80°
∠BCA + ∠ACD = 80°
So, ∠ACD = 50° and,
∠ECD = 50°

Q5: In Figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.
Sol: Since angle which is subtended by an arc at the centre of the circle is double the angle subtended by that arc at any point on the remaining part of the circle.
So, the reflex ∠POR = 2 × ∠PQR
We know the values of angle PQR as 100°
So, ∠POR = 2 × 100° = 200°
∴ ∠POR = 360° – 200° = 160°
Now, in ΔOPR,
OP and OR are the radii of the circle
So, OP = OR
Also, ∠OPR = ∠ORP
Now, we know sum of the angles in a triangle is equal to 180 degrees
So,
∠POR + ∠OPR + ∠ORP = 180°
⇒ ∠OPR + ∠OPR = 180° – 160°
As ∠OPR = ∠ORP
⇒ 2∠OPR = 20°
Thus, ∠OPR = 10°

Q6: In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Sol: Consider this diagram:
Given: In ∆ABC, AD is the angle bisector of ∠A and OD is the perpendicular bisector of BC, intersecting each other at point D.
To Prove: D lies on the circle
Construction: Join OB and OC
Proof:
BC is a chord of the circle.
The perpendicular bisector will pass through centre O of the circumcircle.
∴ OE ⊥ BC & E is the midpoint of BC
Chord BC subtends twice the angle at the centre, as compared to any other point.
BC subtends ∠BAC on the circle & BC subtends ∠BOC on the centre
∴ ∠BAC = 1/2 ∠ BOC
In ∆ BOE and ∆COE,
BE = CE (OD bisects BC)
∠BEO = ∠CEO (Both 90°, as OD ⊥ BC)
OE = OE (Common)
∴ ∆BOE ≅ ∆COE (SAS Congruence rule)
∴ ∠BOE = ∠COE (CPCT)
Now,
∠BOC = ∠BOE + ∠COE
∠BOC = ∠BOE + ∠BOE
∠BOC = 2 ∠BOE …(2)
AD is angle bisector of ∠A
∴ ∠BAC = 2∠BAD
From (1)
∠BAC = 1/2 ∠BOC
2 ∠BAD = 1/2 (2∠BOE)
2 ∠BAD = ∠BOE
∠BAD = 1/2 ∠BOE
BD subtends ∠BOE at centre and half of its angle at Point A.
Hence, BD must be a chord.
∴ D lies on the circle.

Q7: Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6, find the radius of the circle.
Sol:
Here, OM ⊥ AB and ON ⊥ CD. is drawn and OB and OD are joined.
As we know, AB bisects BM as the perpendicular from the centre bisects the chord.
Since AB = 5 so,
BM = AB/2
Similarly, ND = CD/2 = 11/2
Now, let ON be x.
So, OM = 6− x.
Consider ΔMOB,
OB² = OM² + MB²
Or,
OB² = 36 + x² – 12x + 25/4 ……(1)
Consider ΔNOD,
OD² = ON² + ND²
Or,
OD² = x² + 121/4 ……….(2)
We know, OB = OD (radii)
From eq. (1) and eq. (2) we have;
36 + x² -12x + 25/4 = x² + 121/4
12x = 36 + 25/4 – 121/4
12x = (144 + 25 -121)/4
12x = 48/4 = 12
x = 1
Now, from eq. (2) we have,
OD² = 11 + (121/4)
Or OD = (5/2) × √5

Q8: If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Sol: Construction-Consider a trapezium ABCD with AB||CD and BC = AD.
Draw AM ⊥CD and BN ⊥ CD
In ∆AMD and ∆BNC;
AD = BC (Given)
∠AMD = ∠BNC (90°)
AM =BN (perpendiculars between parallel lines)
∆AMD = ∆BNC (By RHS congruency)
∆ADC = ∆BCD (By CPCT rule) …….(i)
∠BAD and ∠ADC are on the same side of transversal AD.
∠BAD + ∠ADC = 180° ……(ii)
∠BAD + ∠BCD = 180° (by equation (i))
Since, the opposite angles are supplementary, therefore, ABCD is a cyclic quadrilateral.

Q9: In Figure, ∠ABC = 69°, ∠ ACB = 31°, find ∠BDC.
Sol:

As we know, angles in the segment of the circle are equal so,
∠BAC = ∠BDC
Now in the In ΔABC, sum of all the interior angles will be 180°
So, ∠ABC + ∠BAC + ∠ACB = 180°
Now, by putting the values,
∠BAC = 180° – 69° – 31°
So, ∠BAC = 80°

Q10: A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.
Sol: First, draw a diagram according to the given statements. The diagram will look as follows.
Here the positions of Ankur, Syed and David are represented as A, B and C respectively. Since they are sitting at equal distances, the triangle ABC will form an equilateral triangle.
AD ⊥ BC is drawn. Now, AD is median of ΔABC and it passes through the centre O.
Also, O is the centroid of the ΔABC. OA is the radius of the triangle.
OA = 2/3 AD
Let the side of a triangle a metres then BD = a/2 m.
Applying Pythagoras theorem in ΔABD,
AB² = BD² + AD²
⇒ AD² = AB² – BD²
⇒ AD² = a² – (a/2)²
⇒ AD² = 3a²/4
⇒ AD = √3a/2
OA = 2/3 AD
⇒ 20 m = 2/3 × √3a/2
⇒ a = 20√3 m
So, the length of the string of the toy is 20√3 m.

Q1: What is a quadrilateral? Mention 6 types of quadrilaterals.
Sol: A quadrilateral is a two-dimensional closed polygon having four sides and four vertices. The sum of its interior angles is 360°. Common types of quadrilaterals include:

• Rectangle
• Square
• Parallelogram
• Rhombus
• Trapezium
• Kite

Q2: Identify the type of quadrilaterals:
(i) The quadrilateral formed by joining the midpoints of consecutive sides of a quadrilateral whose diagonals are perpendicular.
(ii) The quadrilateral formed by joining the midpoints of consecutive sides of a quadrilateral whose diagonals are congruent.
Sol: (i) By Varignon’s theorem, the quadrilateral formed by joining the midpoints of the sides of any quadrilateral is a parallelogram whose adjacent sides are parallel to the diagonals of the original quadrilateral and have lengths equal to half those diagonals. If the original diagonals are perpendicular, the two adjacent sides of the midpoint parallelogram are perpendicular. A parallelogram with adjacent sides perpendicular is a rectangle. Hence the figure is a rectangle.

(ii) If the diagonals of the original quadrilateral are congruent (equal in length), then the corresponding adjacent sides of the midpoint parallelogram are equal in length (each equals half a diagonal). A parallelogram whose adjacent sides are equal is a rhombus. Hence the figure is a rhombus.

Q3: In a rectangle, one diagonal is inclined to one of its sides at 25°. Measure the acute angle between the two diagonals.
Sol: Let θ = 25° be the acute angle made by a diagonal with a side of the rectangle. The two diagonals of a rectangle are symmetric about the centre and make equal angles θ and -θ with the same side. Therefore the acute angle between the two diagonals is 2θ = 2 × 25° = 50°. Hence the acute angle between the diagonals is 50°.

Q4: Prove that the angle bisectors of a parallelogram form a rectangle.
Sol: Let LMNO be a parallelogram. Let the bisectors of ∠L, ∠M, ∠N and ∠O meet pairwise to form quadrilateral PQRS (so each vertex of PQRS is the intersection of bisectors of two consecutive angles of LMNO).

In a parallelogram, consecutive interior angles are supplementary, so ∠L + ∠M = 180°.

When ∠L and ∠M are bisected, their half-angles satisfy (∠L)/2 + (∠M)/2 = 90°.

At the intersection point Q of the bisectors of ∠L and ∠M, the angle ∠PQR equals (∠L)/2 + (∠M)/2 = 90°.

Similarly, each interior angle of PQRS is formed by the sum of half of two consecutive angles of the parallelogram, and each such sum equals 90°.

Hence all four angles of PQRS are right angles, so PQRS is a rectangle. 

Q5: Calculate all the angles of a parallelogram if one of its angles is twice its adjacent angle.

Sol: Let one angle be x and its adjacent angle be 2x. Opposite angles of a parallelogram are equal, so the four angles are x, 2x, x and 2x. Their sum is 360°:

x + 2x + x + 2x = 360°

6x = 360° ⇒ x = 60°.

Thus the angles are 60°, 120°, 60° and 120°.

Q6: The diagonals of which quadrilateral are equal and bisect each other at 90°?
Sol: A square. In a square the diagonals are equal in length, they bisect each other, and they intersect at right angles. A rhombus has diagonals that bisect at 90° but they are not equal unless it is a square; a rectangle has equal diagonals but they are not perpendicular unless it is a square. Therefore the required quadrilateral is a square.

Q7: Find all the angles of a parallelogram if one angle is 80°.

Sol: In a parallelogram opposite angles are equal. If one angle is 80°, the opposite angle is also 80°. Consecutive angles are supplementary, so each adjacent angle is 180° – 80° = 100°. Hence the four angles are 80°, 100°, 80° and 100°.

Q8: Is it possible to draw a quadrilateral whose all angles are obtuse angles?

Sol: No. An obtuse angle is greater than 90°. If all four angles of a quadrilateral were obtuse, their sum would be greater than 4 × 90° = 360°, but the sum of interior angles of any quadrilateral is exactly 360°. Therefore it is not possible to have all four angles obtuse.

Q9: In a trapezium ABCD, AB∥CD. Calculate ∠C and ∠D if ∠A = 55° and ∠B = 70°

Sol: In a trapezium with AB ∥ CD, each pair of interior angles on the same leg are supplementary. Thus ∠A + ∠D = 180° and ∠B + ∠C = 180°.
Given ∠A = 55°, so ∠D = 180° – 55° = 125°.
Given ∠B = 70°, so ∠C = 180° – 70° = 110°.
Therefore ∠C = 110° and ∠D = 125°.

Q10: Calculate all the angles of a quadrilateral if they are in the ratio 2:5:4:1.

Sol: Let the common factor be x. Then the angles are 2x, 5x, 4x and x. Their sum is 360°:
2x + 5x + 4x + x = 360° ⇒ 12x = 360° ⇒ x = 30°.
Therefore the angles are: 2x = 60°, 5x = 150°, 4x = 120°, and x = 30°.

Q1: ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that
(i) ΔABD ≅ ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.
Sol:
As per given in the question,
∠DAB = ∠CBA and AD = BC.
(i) ΔABD and ΔBAC are similar by SAS congruency as
AB = BA (common arm)
∠DAB = ∠CBA and AD = BC (given)
So, triangles ABD and BAC are similar
i.e. ΔABD ≅ ΔBAC. (Hence proved).
(ii) As it is already proved,
ΔABD ≅ ΔBAC
So,
BD = AC (by CPCT)
(iii) Since ΔABD ≅ ΔBAC
So, the angles,
∠ABD = ∠BAC (by CPCT).

Q2: Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A. Show that:
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.Sol:
It is given that the line “l” is the bisector of angle ∠A and the line segments BP and BQ are perpendiculars drawn from l.
(i) ΔAPB and ΔAQB are similar by AAS congruency because;
∠P = ∠Q (both are right angles)
AB = AB (common arm)
∠BAP = ∠BAQ (As line l is the bisector of angle A)
So, ΔAPB ≅ ΔAQB.
(ii) By the rule of CPCT, BP = BQ. So, we can say point B is equidistant from the arms of ∠A.

Q3: In right triangle ABC, right-angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the figure). Show that:
(i) ΔAMC ≅ ΔBMD
(ii) ∠DBC is a right angle.
(iii) ΔDBC ≅ ΔACB
(iv) CM = 1/2 AB
Sol:
It is given that M is the mid-point of the line segment AB, ∠C = 90°, and DM = CM
(i) Consider the triangles ΔAMC and ΔBMD:
AM = BM (Since M is the mid-point)
CM = DM (Given)
∠CMA = ∠DMB (Vertically opposite angles)
So, by SAS congruency criterion, ΔAMC ≅ ΔBMD.

(ii) ∠ACM = ∠BDM (by CPCT)
∴ AC ∥ BD as alternate interior angles are equal.
Now, ∠ACB + ∠DBC = 180° (Since they are co-interiors angles)
⇒ 90° + ∠B = 180°
∴ ∠DBC = 90°

(iii) In ΔDBC and ΔACB,
BC = CB (Common side)
∠ACB = ∠DBC (Both are right angles)
DB = AC (by CPCT)
So, ΔDBC ≅ ΔACB by SAS congruency.

(iv) DC = AB (Since ΔDBC ≅ ΔACB)
⇒ DM = CM = AM = BM (Since M the is mid-point)
So, DM + CM = BM + AM
Hence, CM + CM = AB
⇒ CM = (½) AB

Q4: ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show that ∠BCD is a right angle.Sol:
Given, AB = AC and AD = AB
To prove: ∠BCD is a right angle.
Proof:
Consider ΔABC,
AB = AC (Given)
Also, ∠ACB = ∠ABC (Angles opposite to equal sides)
Now, consider ΔACD,
AD = AC
Also, ∠ADC = ∠ACD (Angles opposite to equal sides)
Now,
In ΔABC,
∠CAB + ∠ACB + ∠ABC = 180°
So, ∠CAB + 2∠ACB = 180°
⇒ ∠CAB = 180° – 2∠ACB — (i)
Similarly in ΔADC,
∠CAD = 180° – 2∠ACD — (ii)
Also,
∠CAB + ∠CAD = 180° (BD is a straight line.)
Adding (i) and (ii) we get,
∠CAB + ∠CAD = 180° – 2∠ACB + 180° – 2∠ACD
⇒ 180° = 360° – 2∠ACB – 2∠ACD
⇒ 2(∠ACB + ∠ACD) = 180°
⇒ ∠BCD = 90°

Q5: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see the figure). Show that:
(i) ΔABM ≅ ΔPQN
(ii) ΔABC ≅ ΔPQR
Sol:
Given;
AB = PQ,
BC = QR and
AM = PN
(i) 1/2 BC = BM and 1/2QR = QN (Since AM and PN are medians)
Also, BC = QR
So, 1/2 BC = 1/2QR
⇒ BM = QN
In ΔABM and ΔPQN,
AM = PN and AB = PQ (Given)
BM = QN (Already proved)
∴ ΔABM ≅ ΔPQN by SSS congruency.

(ii) In ΔABC and ΔPQR,
AB = PQ and BC = QR (Given)
∠ABC = ∠PQR (by CPCT)
So, ΔABC ≅ ΔPQR by SAS congruency.

Q6: AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.
Sol:
Given, AD and BC are two equal perpendiculars to AB.
To prove: CD is the bisector of AB
Proof:
Triangles ΔAOD and ΔBOC are similar by AAS congruency
Since:
(i) ∠A = ∠B (perpendicular angles)
(ii) AD = BC (given)
(iii) ∠AOD = ∠BOC (vertically opposite angles)
∴ ΔAOD ≅ ΔBOC.
So, AO = OB ( by CPCT).
Thus, CD bisects AB (Hence proved).

Q7: AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB. Show that
(i) ΔDAP ≅ ΔEBP
(ii) AD = BE

Sol:
Given, P is the mid-point of line segment AB.
Also, ∠BAD = ∠ABE and ∠EPA = ∠DPB
(i) Given, ∠EPA = ∠DPB
Now, add ∠DPE on both sides,
∠EPA + ∠DPE = ∠DPB + ∠DPE
This implies that angles DPA and EPB are equal
i.e. ∠DPA = ∠EPB
Now, consider the triangles DAP and EBP.
∠DPA = ∠EPB
AP = BP (Since P is the mid-point of the line segment AB)
∠BAD = ∠ABE (given)
So, by ASA congruency criterion,
ΔDAP ≅ ΔEBP.
(ii) By the rule of CPCT,
AD = BE

Q8: ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that these altitudes are equal.Sol:
Given:
(i) BE and CF are altitudes.
(ii) AC = AB
To prove:
BE = CF
Proof:
Triangles ΔAEB and ΔAFC are similar by AAS congruency, since;
∠A = ∠A (common arm)
∠AEB = ∠AFC (both are right angles)
AB = AC (Given)
∴ ΔAEB ≅ ΔAFC
and BE = CF (by CPCT).

Q9: ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the figure). If AD is extended to intersect BC at P, show that
(i) ΔABD ≅ ΔACD
(ii) ΔABP ≅ ΔACP
(iii) AP bisects ∠A as well as ∠D.
(iv) AP is the perpendicular bisector of BC.
Sol:
In the above question, it is given that ΔABC and ΔDBC are two isosceles triangles.
(i) ΔABD and ΔACD are similar by SSS congruency because:
AD = AD (It is the common arm)
AB = AC (Since ΔABC is isosceles)
BD = CD (Since ΔDBC is isosceles)
∴ ΔABD ≅ ΔACD.

(ii) ΔABP and ΔACP are similar as:
AP = AP (common side)
∠PAB = ∠PAC ( by CPCT since ΔABD ≅ ΔACD)
AB = AC (Since ΔABC is isosceles)
So, ΔABP ≅ ΔACP by SAS congruency.

(iii) ∠PAB = ∠PAC by CPCT as ΔABD ≅ ΔACD.
AP bisects ∠A. ………… (1)
Also, ΔBPD and ΔCPD are similar by SSS congruency as
PD = PD (It is the common side)
BD = CD (Since ΔDBC is isosceles.)
BP = CP (by CPCT as ΔABP ≅ ΔACP)
So, ΔBPD ≅ ΔCPD.
Thus, ∠BDP = ∠CDP by CPCT. ……………. (2)
Now by comparing equation (1) and (2) it can be said that AP bisects ∠A as well as ∠D.

(iv) ∠BPD = ∠CPD (by CPCT as ΔBPD ≅ ΔCPD)
and BP = CP — (1)
also,
∠BPD + ∠CPD = 180° (Since BC is a straight line.)
⇒ 2∠BPD = 180°
⇒ ∠BPD = 90° —(2)
Now, from equations (1) and (2), it can be said that
AP is the perpendicular bisector of BC.

Q10: In the Figure, PR > PQ and PS bisect ∠QPR. Prove that ∠PSR > ∠PSQ.
Sol:
Given, PR > PQ and PS bisects ∠QPR
To prove: ∠PSR > ∠PSQ
Proof:
∠QPS = ∠RPS — (1) (PS bisects ∠QPR)
∠PQR > ∠PRQ — (2) (Since PR > PQ as angle opposite to the larger side is always larger)
∠PSR = ∠PQR + ∠QPS — (3) (Since the exterior angle of a triangle equals the sum of opposite interior angles)
∠PSQ = ∠PRQ + ∠RPS — (4) (As the exterior angle of a triangle equals to the sum of opposite interior angles)
By adding (1) and (2)
∠PQR + ∠QPS > ∠PRQ + ∠RPS
Now, from (1), (2), (3) and (4), we get
∠PSR > ∠PSQ.

Q1: In the figure, lines AB and CD intersect at O. If ∠AOC + ∠BOE = 70° and ∠BOD = 40°, find ∠BOE and reflex ∠COE.

Ans:
At point O three adjacent angles on a straight line add to 180°.
So, ∠AOC + ∠BOE + ∠COE = 180°.
Given ∠AOC + ∠BOE = 70°, substitute to get:
70° + ∠COE = 180°
∴ ∠COE = 110°.
Now consider the other straight line through O: ∠COE + ∠BOD + ∠BOE = 180°.
Substitute ∠COE = 110° and ∠BOD = 40°:
110° + 40° + ∠BOE = 180°
∴ ∠BOE = 180° – 150° = 30°.
The reflex angle ∠COE is the larger angle at O corresponding to ∠COE, so reflex ∠COE = 360° – 110° = 250°.
Hence, ∠BOE = 30° and reflex ∠COE = 250°.
Q2: In the Figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = 1/2(∠QOS – ∠POS).

Ans:
POQ is a straight line, so the three adjacent angles on that line sum to 180°:
∠POS + ∠ROS + ∠ROQ = 180°.
Given OR ⟂ PQ, so OR is perpendicular to the straight line POQ. Hence ∠POR = ∠ROQ = 90°. In particular ∠ROQ = 90°.
Substitute ∠ROQ = 90° into the previous equation:
∠POS + ∠ROS + 90° = 180°
∴ ∠POS + ∠ROS = 90°. (1)
Now ∠QOS is formed by ∠QOR + ∠ROS. But ∠QOR = ∠ROQ = 90°, so
∠QOS = 90° + ∠ROS.
Rearrange this to get:
∠QOS – ∠ROS = 90°. (2)
From (1) and (2) we have:
∠POS + ∠ROS = ∠QOS – ∠ROS
Bring like terms together:
2∠ROS + ∠POS = ∠QOS
Therefore ∠ROS = 1/2(∠QOS – ∠POS), as required.
Q3: In the Figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
Ans:
Since AB ∥ CD and GE is a transversal, alternate interior angles are equal. Therefore
∠AGE = ∠GED = 126°.
At point E on line EF, ∠GED is split as ∠GEF + ∠FED. Given EF ⟂ CD, we have ∠FED = 90°.
So, ∠GEF = ∠GED – ∠FED = 126° – 90° = 36°.
Also, ∠FGE and ∠GED are a linear pair (they lie on a straight line through G), so
∠FGE + ∠GED = 180°.
Hence ∠FGE = 180° – 126° = 54°.
Therefore,
∠AGE = 126°, ∠GEF = 36° and ∠FGE = 54°.
Q4: In Fig. 6.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB || CD.
Ans:
Draw normals BE and CF at the points of incidence B and C respectively, so BE ⟂ PQ and CF ⟂ RS. Since PQ ∥ RS, their normals BE and CF are also parallel: BE ∥ CF.
By the law of reflection, angle of incidence = angle of reflection at each mirror. So at B:
∠(AB, BE) = ∠(EB, BC). Label these ∠1 = ∠2. At C:
∠(BC, CF) = ∠(CF, CD). Label these ∠3 = ∠4.
Because BE ∥ CF and BC is a transversal cutting them at B and C, alternate interior angles are equal. Thus ∠2 = ∠3.
Combine equalities: ∠1 = ∠2 and ∠2 = ∠3 and ∠3 = ∠4, so ∠1 = ∠4.
Angles ∠1 and ∠4 are the angles that AB and CD make with the common direction of the mirrors. Since these corresponding angles are equal, AB ∥ CD (alternate interior/corresponding angle test).
Thus AB is parallel to CD.
Q5: In the figure, if AB || CD || EF, PQ || RS, ∠RQD = 25° and ∠CQP = 60°, then find ∠QRS.
Ans:
Given PQ ∥ RS and the transversal through Q and C, angles ∠PQC and ∠BRS are alternate interior angles, so they are equal. Thus ∠BRS = ∠PQC = 60°. (i)
Given AB ∥ CD and the transversal QR, angles ∠DQR and ∠QRA are alternate interior angles, so ∠QRA = ∠DQR = 25°. (ii)
Angles ∠ARS and ∠BRS form a linear pair on line RS, so ∠ARS + ∠BRS = 180°. Using (i):
∴ ∠ARS = 180° – 60° = 120°. (iii)
Now ∠QRS = ∠QRA + ∠ARS. Using (ii) and (iii):
∠QRS = 25° + 120° = 145°.
Hence ∠QRS = 145°.
Q6: In the Figure, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.
Ans:
Angles on a straight line add to 180°, so along the line through P-O-Y we have:
∠POY + a + b = 180°. Given ∠POY = 90°, so a + b = 90°.
Given a : b = 2 : 3, let a = 2x and b = 3x. Then 2x + 3x = 90° ⇒ 5x = 90° ⇒ x = 18°.
Thus a = 2×18° = 36° and b = 3×18° = 54°.
From the diagram b and c form a straight angle on the other line, so b + c = 180°. Substitute b = 54°:
c + 54° = 180° ⇒ c = 126°.
Therefore c = 126°.
Q7: It is given that ∠XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.
Ans:
Since XY is produced to P, XP is a straight line. Therefore ∠XYZ + ∠ZYP = 180°.
Given ∠XYZ = 64°, so ∠ZYP = 180° – 64° = 116°.
Ray YQ bisects ∠ZYP, so each part is half of 116°: ∠ZYQ = ∠QYP = 58°.
Now ∠XYQ = ∠XYZ + ∠ZYQ = 64° + 58° = 122°.
The reflex angle ∠QYP is the larger angle at Y from Q to P around Y: reflex ∠QYP = 360° – (minor ∠QYP). The minor ∠QYP = 58°, so
reflex ∠QYP = 360° – 58° = 302°. (Equivalently, reflex ∠QYP = 180° + ∠XYQ = 180° + 122° = 302°.)
Hence ∠XYQ = 122° and reflex ∠QYP = 302°.
Q8: In the Figure, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
[Hint: Draw a line parallel to ST through point R.]

Ans:
Through R draw a line XY parallel to PQ (and hence parallel to ST).
Angles on the same side of a transversal add to 180°. For transversal QR with PQ ∥ XY:
∠PQR + ∠QRX = 180° ⇒ ∠QRX = 180° – 110° = 70°.
For transversal RS with ST ∥ XY:
∠RST + ∠SRY = 180° ⇒ ∠SRY = 180° – 130° = 50°.
On the straight line XY at R the three adjacent angles satisfy:
∠QRX + ∠QRS + ∠SRY = 180°.
Substitute the known angles: 70° + ∠QRS + 50° = 180° ⇒ ∠QRS = 180° – 120° = 60°.
Thus ∠QRS = 60°.
Q9: In Fig. 6.40, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of Δ XYZ, find ∠OZY and ∠YOZ.

Ans:
Sum of interior angles of ΔXYZ: ∠X + ∠XYZ + ∠XZY = 180°.
Substitute ∠X = 62° and ∠XYZ = 54° to get:
62° + 54° + ∠XZY = 180° ⇒ ∠XZY = 180° – 116° = 64°.
ZO bisects ∠XZY, so ∠OZY = 1/2 × 64° = 32°.
YO bisects ∠XYZ, so ∠OYZ = 1/2 × 54° = 27°.
Now the angles in triangle OYZ sum to 180°:
∠OZY + ∠OYZ + ∠YOZ = 180° ⇒ 32° + 27° + ∠YOZ = 180°.
Therefore ∠YOZ = 180° – 59° = 121°.
Hence ∠OZY = 32° and ∠YOZ = 121°.

Q1: What are the five postulates of Euclid’s Geometry?
Sol: Euclid’s postulates are the basic assumptions on which Euclidean geometry is built. They are:

• A straight line may be drawn from any one point to any other point.
• A terminated straight line can be produced indefinitely in a straight line.
• A circle can be described with any centre and any radius.
• All right angles are equal to one another.
• If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of the angles is less than two right angles.

These five statements serve as the starting rules for classical plane geometry. The fifth postulate is often called the parallel postulate because of its relation to the existence and behaviour of parallel lines.
Q2: If in Q.2, point C is called a mid-point of line segment AB. Prove that every line segment has one and only one mid-point.
Sol:

Let AB be a line segment.
Assume, for contradiction, that P and Q are two distinct mid-points of AB.
Then by definition of midpoint,
AP = PB ………(1)
and AQ = QB ……(2)
Since AP + PB = AB and AQ + QB = AB, both AP and AQ are halves of AB.
From (1), adding AP to both sides gives:
AP + AP = PB + AP (If equals are added to equals, the wholes are equal.)
⇒ 2 AP = AB …(3)
From (2), similarly:
2 AQ = AB …(4)
From (3) and (4):
2 AP = 2 AQ
⇒ AP = AQ
Hence the distances from A to P and from A to Q are equal, so P and Q coincide.
This contradicts the assumption that P and Q are different points.
Therefore every line segment has one and only one mid-point.
Hence proved.
Q3: Does Euclid’s fifth postulate imply the existence of parallel lines? Explain.
Sol:
Yes. Euclid’s fifth postulate implies the existence of parallel lines because it gives a condition under which two straight lines will never meet.

If a straight line falling on two straight lines makes the interior angles on the same side add up to two right angles (180°), then the two lines, when extended indefinitely on that side, do not meet. Such lines are defined as parallel.
Thus, when ∠1 + ∠2 = 180° (or ∠3 + ∠4 = 180° in the figure), the two lines are parallel.
Equivalently, this postulate is closely related to Playfair’s axiom, which states: through a point not on a given line there is exactly one line parallel to the given line. Both statements describe the same geometric idea in different forms.
Q4: If a point C lies between two points A and B such that AC = BC, then prove that AC =1/2 AB. Explain by drawing the figure.
Sol:

Given AC = BC and C lies between A and B, so AB = AC + CB.
Add AC to both sides of AC = BC:
AC + AC = BC + AC
2 AC = BC + AC
But BC + AC = AB (since AC and CB together make AB), therefore:
2 AC = AB
⇒ AC = 1/2 AB.
Thus AC is half of AB, so C is the midpoint of AB.
Q5: In the given figure, if AC = BD, then prove that AB = CD.
Sol:

It is given that AC = BD.
From the figure, AC = AB + BC and BD = BC + CD.
Therefore,
AB + BC = BC + CD (since AC = BD)
Subtract BC from both sides (if equals are subtracted from equals, the remainders are equal):
AB = CD.
Hence proved that AB equals CD.
Q.6: It is known that x + y = 10 and that x = z. Show that z + y = 10.
Sol:
Given:
x + y = 10 …(i)
x = z …(ii)
Substitute z for x in equation (i) (if equals are substituted for equals, the results are equal):
z + y = 10.
Hence z + y = 10, as required.

Q1: Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
(i) x – y/5 – 10 = 0
(ii) -2x+3y = 6
(iii) y – 2 = 0
Sol:
(i) x – y/5 – 10 = 0
The equation x-y/5-10 = 0 can be written as:
(1)x + (-1/5) y + (-10) = 0
Now compare the above equation with ax + by + c = 0
Thus, we get;
a = 1
b = -⅕
c = -10

(ii) -2x + 3y = 6
Re-arranging the given equation, we get,
–2x + 3y – 6 = 0
The equation –2x + 3y – 6 = 0 can be written as,
(–2)x + 3y +(– 6) = 0
Now comparing (–2)x + 3y +(– 6) = 0 with ax + by + c = 0
We get, a = –2
b = 3
c = -6

(iii) y – 2 = 0
The equation y – 2 = 0 can be written as,
0x + 1y + (–2) = 0
Now comparing 0x + 1y + (–2) = 0with ax + by + c = 0
We get, a = 0
b = 1
c = –2

Q2: Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.
Sol:
The given equation is
2x + 3y = k
According to the question, x = 2 and y = 1.
Now, Substituting the values of x and y in the equation 2x + 3y = k,
We get,
⇒ (2 x 2)+ (3 × 1) = k
⇒ 4+3 = k
⇒ 7 = k
⇒ k = 7
The value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k, is 7.

Q3: If the point (3, 4) lies on the graph of the equation 3y = ax + 7, find the value of a.
Sol:
The given equation is
3y = ax + 7
According to the question, x = 3 and y = 4
Now, Substituting the values of x and y in the equation 3y = ax + 7,
We get,
(3×4) = (ax3) + 7
⇒ 12 = 3a+7
⇒ 3a = 12–7
⇒ 3a = 5
⇒ a = 5/3
The value of a, if the point (3, 4) lies on the graph of the equation 3y = ax + 7 is 5/3.

Q4: Draw the graph of the linear equation 3x + 4y = 6. At what points, the graph cuts X and Y-axis?
Sol: Given equation,
3x + 4y = 6.
We need at least 2 points on the graph to draw the graph of this equation,
Thus, the points the graph cuts
(i) x-axis
Since the point is on the x-axis, we have y = 0.
Substituting y = 0 in the equation, 3x + 4y = 6,
We get,
3x + 4×0 = 6
⇒ 3x = 6
⇒ x = 2
Hence, the point at which the graph cuts x-axis = (2, 0).

(ii) y-axis
Since the point is on the y-axis, we have, x = 0.
Substituting x = 0 in the equation, 3x + 4y = 6,
We get,
3×0 + 4y = 6
⇒ 4y = 6
⇒ y = 6/4
⇒ y = 3/2
⇒ y = 1.5
Hence, the point at which the graph cuts y-axis = (0, 1.5).
Plotting the points (0, 1.5) and (2, 0) on the graph.

Q5: Write four solutions for each of the following equations:
(i) 2x + y = 7
(ii) πx + y = 9
Sol: (i) 2x + y = 7
To find the four solutions of 2x + y = 7 we substitute different values for x and y
Let x = 0
Then,
2x + y = 7
(2×0)+y = 7
y = 7
(0,7)
Let x = 1
Then,
2x + y = 7
(2×1)+y = 7
2+y = 7
y = 7 – 2
y = 5
(1,5)
Let y = 1
Then,
2x + y = 7
2x+ 1 = 7
2x = 7 – 1
2x = 6
x = 3
(3,1)
Let x = 2
Then,
2x + y = 7
2(2)+y = 7
4+y = 7
y = 7 – 4
y = 3
(2,3)
The solutions are (0, 7), (1,5), (3,1), (2,3)

(ii) πx + y = 9
To find the four solutions of πx + y = 9 we substitute different values for x and y
Let x = 0
Then,
πx + y = 9
(π × 0)+y = 9
y = 9
(0,9)
Let x = 1
Then,
πx + y = 9
(π×1)+y = 9
π+y = 9
y = 9-π
(1,9-π)
Let y = 0
Then,
πx + y = 9
πx +0 = 9
πx = 9
x =9/π
(9/π,0)
Let x = -1
Then,
πx + y = 9
(π(-1))+y = 9
-π + y = 9
y = 9+π
(-1,9+π)
The solutions are (0,9), (1,9-π),(9/π,0),(-1,9+π)

Q6: Draw the graph of each of the following linear equations in two variables:
(i)y = 3x
(ii) 3 = 2x + y
Sol: (i) y = 3x
To draw a graph of linear equations in two variables, let us find out the points to plot.
To find out the points, we have to find the values for which x and y satisfies the given equation.
Here,
y=3x
Substituting the values for x,
When x = 0,
y = 3x
y = 3(0)
⇒ y = 0
When x = 1,
y = 3x
y = 3(1)
⇒ y = 3
The points to be plotted are (0, 0) and (1, 3)

(ii) 3 = 2x + y
To draw a graph of linear equations in two variables, let us find out the points to plot.
To find out the points, we have to find the values for which x and y satisfies the given equation.
Here,
3 = 2x + y
Substituting the values for x,
When x = 0,
3 = 2x + y
⇒ 3 = 2(0) + y
⇒ 3 = 0 + y
⇒ y = 3
When x = 1,
3 = 2x + y
⇒ 3 = 2(1) + y
⇒ 3 = 2 + y
⇒ y = 3 – 2
⇒ y = 1
The points to be plotted are (0, 3) and (1, 1)

Q7: Show that the points A (1, 2), B ( – 1, – 16) and C (0, – 7) lie on the graph of the linear equation y = 9x – 7.
Sol:
We have the equation,
y = 9x – 7
For A (1, 2),
Substituting (x,y) = (1, 2),
We get,
2 = 9(1) – 7
2 = 9 – 7
2 = 2
For B (–1, –16),
Substituting (x,y) = (–1, –16),
We get,
–16 = 9(–1) – 7
-16 = – 9 – 7
-16 = – 16
For C (0, –7),
Substituting (x,y) = (0, –7),
We get,
– 7 = 9(0) – 7
-7 = 0 – 7
-7 = – 7
Hence, the points A (1, 2), B (–1, –16) and C (0, –7) satisfy the line y = 9x – 7.
Thus, A (1, 2), B (–1, –16) and C (0, –7) are solutions of the linear equation y = 9x – 7
Therefore, the points A (1, 2), B (–1, –16), C (0, –7) lie on the graph of linear equation y = 9x – 7.