Std 10 Mathematics Part 2 · Chapter 7

7. Mensuration

Solutions to Practice Sets 7.1 to 7.4 and Problem Set 7, with figures and full working. Maharashtra State Board (Balbharati), Std 10 Mathematics Part 2 (Geometry).

Practice Set 7.1

1Find the volume of a cone if the radius of its base is 1.5 cm and its perpendicular height is 5 cm.

Volume of cone = 13 πr²h

= 13 × 3.14 × 1.5 × 1.5 × 5 (taking π = 3.14)

= 3.14 × 0.75 × 5 = 11.775

∴ Volume of the cone = 11.775 cm³

2Find the volume of a sphere of diameter 6 cm.

Radius r = 3 cm

Volume of sphere = 43 πr³

= 43 × 3.14 × 3 × 3 × 3 (taking π = 3.14)

= 4 × 3.14 × 9 = 113.04

∴ Volume of the sphere = 113.04 cm³

3Find the total surface area of a cylinder if the radius of its base is 5 cm and height is 40 cm.

Total surface area of cylinder = 2πr(r + h)

= 2 × 3.14 × 5 × (5 + 40) (taking π = 3.14)

= 31.4 × 45 = 1413

∴ Total surface area = 1413 cm²

4Find the surface area of a sphere of radius 7 cm.

Surface area of sphere = 4πr²

= 4 × 227 × 7 × 7

= 4 × 22 × 7 = 616

∴ Surface area = 616 cm²

5The dimensions of a cuboid are 44 cm, 21 cm, 12 cm. It is melted and a cone of height 24 cm is made. Find the radius of its base.

The cuboid is melted to make the cone, so their volumes are equal.

∴ 13 πr²h = 44 × 21 × 12

∴ 13 × 227 × r² × 24 = 44 × 21 × 12

∴ r² = 44 × 21 × 12 × 3 × 722 × 24 = 441

∴ r = 21

∴ Radius of the base of the cone = 21 cm

6Observe the measures of pots in figure 7.8 and 7.9. How many jugs of water can the cylindrical pot hold?

Jug: radius 3.5 cm, height 10 cm. Pot: radius 7 cm, height 10 cm.

Volume of conical jug = 13 π × 3.5 × 3.5 × 10

Volume of cylindrical pot = π × 7 × 7 × 10

Number of jugs = Volume of potVolume of jug = π × 7 × 7 × 1013 × π × 3.5 × 3.5 × 10

= 3 × 7 × 73.5 × 3.5 = 3 × 2 × 2 = 12

∴ The cylindrical pot can hold 12 jugs of water.

7A cylinder and a cone have equal bases. The height of the cylinder is 3 cm and the area of its base is 100 cm². The cone is placed upon the cylinder. Volume of the solid figure so formed is 500 cm³. Find the total height of the figure.

Fig. 7.10

Volume of cylinder = base area × height = 100 × 3 = 300 cm³

Volume of cone = 500 − 300 = 200 cm³

Volume of cone = 13 × base area × h

∴ 200 = 13 × 100 × h

∴ h = 6 cm

Total height = 3 + 6 = 9 cm

∴ Total height of the figure = 9 cm

8In figure 7.11, a toy made from a hemisphere, a cylinder and a cone is shown. Find the total area of the toy.

Fig. 7.11

Radius of hemisphere, cylinder and cone, r = 3 cm. Height of cone = 4 cm.

Height of cylinder = 40 − 3 − 4 = 33 cm

Slant height of cone, l = √4² + 3² = √25 = 5 cm

Total area = curved area of hemisphere + curved area of cylinder + curved area of cone

= 2πr² + 2πrh + πrl

= πr(2r + 2h + l)

= 227 × 3 × (6 + 66 + 5)

= 227 × 3 × 77 = 726

∴ Total area of the toy = 726 cm²

9In the figure 7.12, a cylindrical wrapper of flat tablets is shown. The radius of a tablet is 7 mm and its thickness is 5 mm. How many such tablets are wrapped in the wrapper?

Fig. 7.12

Radius of the wrapper = 14 ÷ 2 = 7 mm = radius of a tablet.

Length of the wrapper = 10 cm = 100 mm

Since the radii are equal, number of tablets = length of wrapperthickness of tablet = 1005 = 20

∴ 20 tablets are wrapped in the wrapper.

10Figure 7.13 shows a toy. Its lower part is a hemisphere and the upper part is a cone. Find the volume and the surface area of the toy from the measures shown in the figure. (π = 3.14)

Radius 3 cm, height of cone 4 cm

Radius r = 3 cm, height of cone h = 4 cm.

Volume = volume of hemisphere + volume of cone

= 23 πr³ + 13 πr²h = 13 πr²(2r + h)

= 13 × 3.14 × 9 × (6 + 4) = 3.14 × 3 × 10 = 94.2

Slant height l = √4² + 3² = 5 cm

Surface area = 2πr² + πrl = πr(2r + l)

= 3.14 × 3 × (6 + 5) = 3.14 × 33 = 103.62

∴ Volume = 94.2 cm³ and surface area = 103.62 cm²

The figure is read as radius 3 cm and height of the cone 4 cm.

11Find the surface area and the volume of a beach ball shown in the figure.

Diameter 42 cm

Radius r = 21 cm

Surface area = 4πr² = 4 × 227 × 21 × 21 = 5544

Volume = 43 πr³ = 43 × 227 × 21 × 21 × 21 = 38808

∴ Surface area = 5544 cm² and volume = 38808 cm³

12As shown in the figure, a cylindrical glass contains water. A metal sphere of diameter 2 cm is immersed in it. Find the volume of the water.

Fig. 7.15

Radius of glass = 7 cm, height = 30 cm. Radius of sphere = 1 cm.

Volume of glass = πr²h = 227 × 7 × 7 × 30 = 4620 cm³

Volume of sphere = 43 π × 1³ = 43 × 227 = 8821 ≈ 4.19 cm³

Volume of water = 4620 − 4.19 = 4615.81

∴ Volume of the water = 4615.81 cm³ (approx.)

The glass is taken as full up to 30 cm, with the sphere inside it.

Practice Set 7.2

1The radii of two circular ends of frustum shape bucket are 14 cm and 7 cm. Height of the bucket is 30 cm. How many litres of water it can hold? (1 litre = 1000 cm³)

Volume of frustum = 13 πh(r₁² + r₂² + r₁r₂)

= 13 × 227 × 30 × (196 + 49 + 98)

= 13 × 227 × 30 × 343

= 10 × 22 × 49 = 10780 cm³

= 107801000 litres = 10.78 litres

∴ The bucket can hold 10.78 litres of water.

2The radii of ends of a frustum are 14 cm and 6 cm respectively and its height is 6 cm. Find its i) curved surface area ii) total surface area iii) volume (π = 3.14)

Slant height l = √h² + (r₁ − r₂)² = √6² + 8² = √100 = 10 cm

i) Curved surface area

= π(r₁ + r₂)l = 3.14 × (14 + 6) × 10 = 628

∴ 628 cm²

ii) Total surface area

= π(r₁ + r₂)l + πr₁² + πr₂²

= 628 + 3.14 × (196 + 36) = 628 + 728.48 = 1356.48

∴ 1356.48 cm²

iii) Volume

Volume of frustum = 13 πh(r₁² + r₂² + r₁r₂)

= 13 × 3.14 × 6 × (196 + 36 + 84)

= 2 × 3.14 × 316 = 1984.48

∴ 1984.48 cm³

3The circumferences of circular faces of a frustum are 132 cm and 88 cm and its height is 24 cm. To find the curved surface area of the frustum complete the following activity. (π = 227)

Fig. 7.23

circumference₁ = 2πr₁ = 132

r₁ = 1322π = 21 cm

circumference₂ = 2πr₂ = 88

r₂ = 882π = 14 cm

slant height of frustum, l = √h² + (r₁ − r₂)²

= √24² + 7²

= 25 cm

curved surface area of the frustum = π(r₁ + r₂)l

= π × 35 × 25

= 2750 sq.cm.

Practice Set 7.3

1Radius of a circle is 10 cm. Measure of an arc of the circle is 54°. Find the area of the sector associated with the arc. (π = 3.14)

Area of sector = θ360 × πr²

= 54360 × 3.14 × 10 × 10

= 320 × 314 = 47.1

∴ Area of the sector = 47.1 cm²

2Measure of an arc of a circle is 80° and its radius is 18 cm. Find the length of the arc. (π = 3.14)

Length of arc = θ360 × 2πr

= 80360 × 2 × 3.14 × 18

= 29 × 113.04 = 25.12

∴ Length of the arc = 25.12 cm

The textbook prints “measure of an arc is 80 cm”. The measure of an arc is an angle, so it is taken as 80°.

3Radius of a sector of a circle is 3.5 cm and length of its arc is 2.2 cm. Find the area of the sector.

Area of sector = length of arc × radius2

= 2.2 × 3.52 = 3.85

∴ Area of the sector = 3.85 cm²

4Radius of a circle is 10 cm. Area of a sector of the circle is 100 cm². Find the area of its corresponding major sector. (π = 3.14)

Area of circle = πr² = 3.14 × 10 × 10 = 314 cm²

Area of major sector = area of circle − area of minor sector

= 314 − 100 = 214

∴ Area of the major sector = 214 cm²

5Area of a sector of a circle of radius 15 cm is 30 cm². Find the length of the arc of the sector.

Area of sector = length of arc × radius2

∴ 30 = l × 152

∴ l = 6015 = 4

∴ Length of the arc = 4 cm

6In the figure 7.31, radius of the circle is 7 cm and m(arc MBN) = 60°, find (1) Area of the circle (2) A(O − MBN) (3) A(O − MCN).

m(arc MBN) = 60°

(1)

Area of circle = πr² = 227 × 7 × 7 = 154 cm²

(2)

A(O − MBN) = 60360 × 154 = 1546 = 25.67 cm² (approx.)

(3)

A(O − MCN) = area of circle − A(O − MBN) = 154 − 25.67 = 128.33 cm² (approx.)

The question gives m(arc MBN) = 60°, but Fig. 7.31 shows 68°. The answers use 60°, as stated in the question.

7In figure 7.32, radius of circle is 3.4 cm and perimeter of sector P − ABC is 12.8 cm. Find A(P − ABC).

Fig. 7.32

Perimeter of sector = 2r + length of arc

∴ 12.8 = 2 × 3.4 + l

∴ l = 12.8 − 6.8 = 6 cm

A(P − ABC) = l × r2 = 6 × 3.42 = 10.2

∴ A(P − ABC) = 10.2 cm²

8In figure 7.33, O is the centre of the sector. ∠ ROQ = ∠ MON = 60°. OR = 7 cm, and OM = 21 cm. Find the lengths of arc RXQ and arc MYN. (π = 227)

Fig. 7.33

l(arc RXQ) = 60360 × 2 × 227 × 7 = 446 = 223 ≈ 7.33 cm

l(arc MYN) = 60360 × 2 × 227 × 21 = 1326 = 22 cm

∴ l(arc RXQ) = 223 cm and l(arc MYN) = 22 cm

9In figure 7.34, if A(P − ABC) = 154 cm², radius of the circle is 14 cm, find (1) ∠ APC (2) l(arc ABC).

Fig. 7.34

(1)

Area of sector = θ360 × πr²

∴ 154 = θ360 × 227 × 14 × 14 = θ360 × 616

∴ θ = 154 × 360616 = 90°

∴ ∠ APC = 90°

(2)

Length of arc = θ360 × 2πr = 90360 × 2 × 227 × 14 = 22

∴ l(arc ABC) = 22 cm

10Radius of a sector of a circle is 7 cm. If measure of arc of the sector is (1) 30° (2) 210° (3) three right angles; find the area of the sector in each case.

Area of circle = 227 × 7 × 7 = 154 cm². Area of sector = θ360 × πr² = θ360 × 154

(1) θ = 30°

Area = 30360 × 154 = 15412 = 12.83 cm² (approx.)

(2) θ = 210°

Area = 210360 × 154 = 712 × 154 = 89.83 cm² (approx.)

(3) θ = 3 × 90° = 270°

Area = 270360 × 154 = 34 × 154 = 115.5 cm²

11The area of a minor sector of a circle is 3.85 cm² and the measure of its central angle is 36°. Find the radius of the circle.

Area of sector = θ360 × πr²

∴ 3.85 = 36360 × 227 × r²

∴ 3.85 = 110 × 227 × r²

∴ r² = 3.85 × 10 × 722 = 12.25

∴ r = 3.5

∴ Radius of the circle = 3.5 cm

12In figure 7.35, □ PQRS is a rectangle. If PQ = 14 cm, QR = 21 cm, find the areas of the parts x, y and z.

Fig. 7.35

Part x is a sector with centre Q, radius 14 cm and angle 90°.

Area of x = 90360 × 227 × 14 × 14 = 154 cm²

Part y is a sector with centre R, radius 21 − 14 = 7 cm and angle 90°.

Area of y = 90360 × 227 × 7 × 7 = 38.5 cm²

Area of rectangle = 14 × 21 = 294 cm²

Area of z = 294 − 154 − 38.5 = 101.5 cm²

∴ x = 154 cm², y = 38.5 cm², z = 101.5 cm²

13Δ LMN is an equilateral triangle. LM = 14 cm. As shown in figure, three sectors are drawn with vertices as centres and radius 7 cm. Find, (1) A(Δ LMN) (2) Area of any one of the sectors. (3) Total area of all the three sectors. (4) Area of the shaded region.

Fig. 7.36: the shaded region is the pink part in the middle.

(1)

A(Δ LMN) = √34 × 14² = 49√3 = 49 × 1.73 = 84.77 cm² (taking √3 = 1.73)

(2)

Each angle of the triangle is 60°. Area of one sector = 60360 × 227 × 7 × 7 = 773 = 25.67 cm² (approx.)

(3)

Total area of three sectors = 3 × 773 = 77 cm²

(4)

Area of shaded region = A(Δ LMN) − total area of sectors = 84.77 − 77 = 7.77 cm²

Practice Set 7.4

1In figure 7.43, A is the centre of the circle. ∠ ABC = 45° and AC = 7√2 cm. Find the area of segment BXC.

Fig. 7.43

AB = AC = 7√2 cm (radii), so ∠ ACB = ∠ ABC = 45°.

∴ ∠ BAC = 180° − 45° − 45° = 90°

A(A − BXC) = 90360 × 227 × (7√2)² = 14 × 227 × 98 = 77 cm²

A(Δ BAC) = 12 × AB × AC = 12 × 7√2 × 7√2 = 49 cm²

A(segment BXC) = A(A − BXC) − A(Δ BAC) = 77 − 49 = 28

∴ Area of segment BXC = 28 cm² (taking π = 227)

2In the figure 7.44, O is the centre of the circle. m(arc PQR) = 60°, OP = 10 cm. Find the area of the shaded region. (π = 3.14, √3 = 1.73)

Fig. 7.44

θ = 60°, r = 10 cm

A(O − PQR) = 60360 × 3.14 × 100 = 52.33 cm² (approx.)

Δ OPR is equilateral (OP = OR and ∠ POR = 60°).

A(Δ OPR) = √34 × 10² = 1.734 × 100 = 43.25 cm²

Area of shaded region = 52.33 − 43.25 = 9.08

∴ Area of the shaded region = 9.08 cm² (approx.)

3In the figure 7.45, if A is the centre of the circle. ∠ PAR = 30°, AP = 7.5, find the area of the segment PQR. (π = 3.14)

Fig. 7.45

Area of segment = r²(πθ360 − sin θ2)

= 7.5² (3.14 × 30360 − sin 30°2)

= 56.25 (0.2617 − 0.25)

= 56.25 × 0.0117 ≈ 0.66

Or: A(A − PQR) = 30360 × 3.14 × 56.25 = 14.72 and A(Δ APR) = 12 × 56.25 × sin 30° = 14.06

∴ Area of segment PQR = 14.72 − 14.06 = 0.66 sq. units (approx.)

4In the figure 7.46, if O is the centre of the circle, PQ is a chord. ∠ POQ = 90°, area of shaded region is 114 cm², find the radius of the circle. (π = 3.14)

Fig. 7.46

Area of shaded segment = A(O − PRQ) − A(Δ POQ)

∴ 114 = 90360 × 3.14 × r² − 12 × r²

∴ 114 = r²(0.785 − 0.5) = 0.285r²

∴ r² = 1140.285 = 400

∴ r = 20

∴ Radius of the circle = 20 cm

5A chord PQ of a circle with radius 15 cm subtends an angle of 60° with the centre of the circle. Find the area of the minor as well as the major segment. (π = 3.14, √3 = 1.73)

Chord PQ subtends 60° at the centre O.

θ = 60°, r = 15 cm

A(O − PQ) = 60360 × 3.14 × 225 = 117.75 cm²

Δ OPQ is equilateral. A(Δ OPQ) = √34 × 225 = 1.73 × 2254 = 97.3125 cm²

A(minor segment) = 117.75 − 97.3125 = 20.4375 ≈ 20.44 cm²

A(circle) = 3.14 × 225 = 706.5 cm²

A(major segment) = 706.5 − 20.4375 = 686.0625 ≈ 686.06 cm²

∴ Minor segment = 20.44 cm² and major segment = 686.06 cm² (approx.)

Problem Set 7

1Choose the correct alternative answer for each of the following questions.

(1) The ratio of circumference and area of a circle is 2 : 7. Find its circumference. — (A) 14π
2πrπr² = 2r = 27, so r = 7 and circumference = 2π × 7 = 14π.
(2) If measure of an arc of a circle is 160° and its length is 44 cm, find the circumference of the circle. — (D) 99 cm
160360 × circumference = 44, so circumference = 44 × 94 = 99 cm.
(3) Find the perimeter of a sector of a circle if its measure is 90° and radius is 7 cm. — (B) 25 cm
Arc = 14 × 2 × 227 × 7 = 11 cm. Perimeter = 7 + 7 + 11 = 25 cm.
(4) Find the curved surface area of a cone of radius 7 cm and height 24 cm. — (B) 550 cm²
l = √49 + 576 = 25. Curved surface area = 227 × 7 × 25 = 550 cm².
(5) The curved surface area of a cylinder is 440 cm² and its radius is 5 cm. Find its height. — (A) 44π cm
2π × 5 × h = 440, so h = 44π.
(6) A cone was melted and cast into a cylinder of the same radius as that of the base of the cone. If the height of the cylinder is 5 cm, find the height of the cone. — (A) 15 cm
13πr²H = πr² × 5, so H = 15 cm.
(7) Find the volume of a cube of side 0.01 cm. — (D) 0.000001 cm³
0.01³ = 0.000001
(8) Find the side of a cube of volume 1 m³. — (C) 100 cm
Side = 1 m = 100 cm.

2A washing tub in the shape of a frustum of a cone has height 21 cm. The radii of the circular top and bottom are 20 cm and 15 cm respectively. What is the capacity of the tub? (π = 227)

Volume of frustum = 13 πh(r₁² + r₂² + r₁r₂)

= 13 × 227 × 21 × (400 + 225 + 300)

= 22 × 925 = 20350 cm³ = 20.35 litres

∴ Capacity of the tub = 20350 cm³, that is 20.35 litres

3★Some plastic balls of radius 1 cm were melted and cast into a tube. The thickness, length and outer radius of the tube were 2 cm, 90 cm and 30 cm respectively. How many balls were melted to make the tube?

Outer radius R = 30 cm, inner radius r = 30 − 2 = 28 cm, length h = 90 cm.

Volume of tube = π(R² − r²)h = π(900 − 784) × 90 = 10440π cm³

Volume of one ball = 43 π × 1³ = 43π cm³

Number of balls = 10440π43π = 10440 × 34 = 7830

∴ 7830 balls were melted.

4A metal parallelopiped of measures 16 cm × 11 cm × 10 cm was melted to make coins. How many coins were made if the thickness and diameter of each coin was 2 mm and 2 cm respectively?

Volume of parallelopiped = 16 × 11 × 10 = 1760 cm³

Each coin is a cylinder with r = 1 cm and h = 2 mm = 0.2 cm.

Volume of one coin = πr²h = 227 × 1 × 1 × 0.2 = 4.47 cm³

Number of coins = 1760 ÷ 4.47 = 1760 × 74.4 = 2800

∴ 2800 coins were made.

5The diameter and length of a roller is 120 cm and 84 cm respectively. To level the ground, 200 rotations of the roller are required. Find the expenditure to level the ground at the rate of Rs. 10 per sq.m.

Radius = 60 cm, length h = 84 cm.

Area levelled in one rotation = curved surface area = 2πrh = 2 × 227 × 60 × 84 = 31680 cm² = 3.168 m²

Area levelled in 200 rotations = 3.168 × 200 = 633.6 m²

Expenditure = 633.6 × 10 = Rs. 6336

∴ The expenditure is Rs. 6336

6The diameter and thickness of a hollow metal sphere are 12 cm and 0.01 m respectively. The density of the metal is 8.88 gm per cm³. Find the outer surface area and mass of the sphere.

Outer radius R = 6 cm. Thickness = 0.01 m = 1 cm, so inner radius r = 5 cm.

Outer surface area = 4πR² = 4 × 3.14 × 36 = 452.16 cm² (taking π = 3.14)

Volume of metal = 43 π(R³ − r³) = 43 × 3.14 × (216 − 125) = 43 × 3.14 × 91 ≈ 380.99 cm³

Mass = volume × density = 380.99 × 8.88 ≈ 3383.16 g

∴ Outer surface area = 452.16 cm² and mass ≈ 3383.16 g (about 3.38 kg)

7A cylindrical bucket of diameter 28 cm and height 20 cm was full of sand. When the sand in the bucket was poured on the ground, the sand got converted into a shape of a cone. If the height of the cone was 14 cm, what was the base area of the cone?

Volume of sand = volume of bucket = πr²h = 227 × 14 × 14 × 20 = 12320 cm³

Volume of cone = 13 × base area × height

∴ 12320 = 13 × base area × 14

∴ Base area = 12320 × 314 = 2640

∴ Base area of the cone = 2640 cm²

8The radius of a metallic sphere is 9 cm. It was melted to make a wire of diameter 4 mm. Find the length of the wire.

Radius of wire = 2 mm = 0.2 cm. Let the length of the wire be h.

Volume of wire = volume of sphere

∴ π × 0.2 × 0.2 × h = 43 π × 9 × 9 × 9

∴ 0.04h = 972

∴ h = 24300 cm = 243 m

∴ Length of the wire = 243 m

9The area of a sector of a circle of 6 cm radius is 15π sq.cm. Find the measure of the arc and length of the arc corresponding to the sector.

Area of sector = θ360 × πr²

∴ 15π = θ360 × π × 36

∴ θ = 15 × 36036 = 150°

Length of arc = θ360 × 2πr = 150360 × 2π × 6 = 5π cm

∴ Measure of the arc = 150° and length of the arc = 5π cm (= 15.7 cm, taking π = 3.14)

10In the figure 7.47, seg AB is a chord of a circle with centre P. If PA = 8 cm and distance of chord AB from the centre P is 4 cm, find the area of the shaded portion. (π = 3.14, √3 = 1.73)

PM ⊥ AB, PM = 4 cm

Let PM ⊥ AB. In Δ PMA, ∠ PMA = 90°, PA = 8 and PM = 4 = 12 PA.

∴ ∠ PAM = 30° and ∠ APM = 60° … 30° − 60° − 90° theorem

Similarly ∠ BPM = 60°, so ∠ APB = 120°.

AM = √32 × 8 = 4√3, so AB = 8√3 cm

A(P − AB) = 120360 × 3.14 × 64 = 66.99 cm² (approx.)

A(Δ APB) = 12 × AB × PM = 12 × 8√3 × 4 = 16√3 = 16 × 1.73 = 27.68 cm²

Area of shaded portion = 66.99 − 27.68 = 39.31

∴ Area of the shaded portion = 39.31 cm² (approx.)

11In the figure 7.48, square ABCD is inscribed in the sector A − PCQ. The radius of sector C − BXD is 20 cm. Complete the following activity to find the area of shaded region.

Fig. 7.48

Side of square ABCD = radius of sector C − BXD = 20 cm

Area of square = (side)² = 20² = 400 …… (I)

Area of shaded region inside the square = Area of square ABCD − Area of sector C − BXD

= 400 − θ360 × πr²

= 400 − 90360 × 3.141 × 4001

= 400 − 314 = 86 sq.cm

Radius of bigger sector = length of diagonal of square ABCD = 20√2

Area of the shaded regions outside the square = A(A − PCQ) − A(□ ABCD)

= (θ360 × π × r²) − 20²

= 90360 × 3.14 × (20√2)² − (20)²

= 628 − 400 = 228 sq.cm

∴ total area of the shaded region = 86 + 228 = 314 sq.cm.

12In the figure 7.49, two circles with centres O and P are touching internally at point A. If BQ = 9, DE = 5, complete the following activity to find the radii of the circles.

Fig. 7.49

Let the radius of the bigger circle be R and that of smaller circle be r.

OA, OB, OC and OD are the radii of the bigger circle. ∴ OA = OB = OC = OD = R

PQ = PA = r

OQ = OB − BQ = R − 9

OE = OD − DE = R − 5

As the chords QA and EF of the circle with centre P intersect in the interior of the circle, by the property of internal division of two chords of a circle,

OQ × OA = OE × OF

R − 9 × R = R − 5 × R − 5 … OE = OF

R² − 9R = R² − 10R + 25

R = 25

AQ = 2r = AB − BQ

2r = 50 − 9 = 41

r = 412 = 20.5

∴ Radius of bigger circle = 25 units and radius of smaller circle = 20.5 units.

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