CBSE Class 10 Mathematics Area Related To Circle Worksheet Set 05

Question. The wheel of the engine of a train is \( 14 \frac{2}{7} \) m in circumference makes 7 revolutions in 10 seconds. The speed of the train is
(a) 18 km/hr
(b) 48 km/hr
(c) 36 km/hr
(d) 24 km/hr
Answer: (c) 36 km/hr
Explanation: Given: Circumference of wheel = \( 14 \frac{2}{7} = \frac{100}{7} \) m,
No. of revolutions in 10 seconds = 7
Now, No. of Revolutions = \( \frac{\text{Total distance}}{\text{Circumference of wheel}} \)

\( \implies 7 = \frac{\text{Total distance} \times 7}{100} \)

\( \implies \) Total distance in 10 second = 100 m
\( \therefore \) Distance in 1 hour = \( \frac{100}{10} \times 3600 = 36000 \) m = 36 km
\( \therefore \) Speed = 36 km/hr

Question. If the perimeter of a circle is equal to that of a square, then the ratio of their areas is ______
(a) 22 : 7
(b) 14 : 11
(c) 11 : 14
(d) 7 : 22
Answer: (b) 14 : 11
Explanation: Let the radius of the circle be r and side of the square be a. Then, according to question,
\( 2\pi r = 4a \)

\( \implies a = \frac{2\pi r}{4} = \frac{\pi r}{2} \)
Now, ratio of their areas,
\( \frac{\pi r^2}{a^2} = \frac{\pi r^2}{(\frac{\pi r}{2})^2} \)
\( = \frac{\pi r^2 \times 4}{\pi^2 r^2} \)
\( = \frac{14}{11} \)

\( \implies \pi r^2 : a^2 = 14 : 11 \)

Question. The circumference of a circle whose diameter is 4.2cm is
(a) 13.2cm
(b) 4.2 cm
(c) 22 cm
(d) 11 cm
Answer: (a) 13.2cm
Explanation: Given: Diameter (d) = 4.2 cm
\( \therefore \) Circumference = \( \pi d = \frac{22}{7} \times 4.2 = 13.2 \) cm

Question. The area of a ring having ‘R’ as outer radius and ‘r’ as inner radius is
(a) \( \pi(R + r) \)
(b) \( \pi(R^2 - r^2) \)
(c) \( \pi(R^2 + r^2) \)
(d) \( \pi(R - r) \)
Answer: (b) \( \pi(R^2 - r^2) \)
Explanation: The area of a ring having ‘R’ as outer radius and ‘r’ as inner radius is \( \pi R^2 - \pi r^2 = \pi(R^2 - r^2) \)

Question. If the area of a circle is ‘A’, radius of the circle is ‘r’ and its circumference is ‘C’, then
(a) rC = 2A
(b) \( AC = \frac{r^2}{4} \)
(c) \( \frac{C}{A} = \frac{r}{2} \)
(d) \( \frac{A}{r} = C \)
Answer: (a) rC = 2A
Explanation: Here, Area of circle (A) = \( \pi r^2 \) and Circumference of circle (C) = \( 2\pi r \)

\( \implies rC = r(2\pi r) = 2\pi r^2 = 2A \)

\( \implies rC = 2A \)

Question. Find the area (in cm\(^2\)) of the circle that can be inscribed in a square of side 8 cm.
Answer: Side of square = 8 cm
Side of square = diameter of circle = 8 cm
\( \therefore \) Radius of circle, \( r = \frac{8}{2} = 4 \)cm
Area of circle = \( \pi r^2 \)
\( = \pi(4)^2 \)
\( = \pi \times 4 \times 4 \)
\( = 16\pi \) cm\(^2\)
So, Area of circle is \( 16\pi \) cm\(^2\).

Question. Find the area of the sector of a circle having radius 6 cm and of angle 30°. [Take \( \pi = 3.14 \)]
Answer: Radius of a circle = r = 6 cm
Central angle = \( \theta = 30^\circ \)
\( \therefore \) Area of the sector = \( \frac{\pi r^2 \theta}{360} \)
\( = \left( \frac{3.14 \times 6 \times 6 \times 30^\circ}{360^\circ} \right) \text{ cm}^2 \)
= 9.42 cm\(^2\)

Question. What is the perimeter of the sector with radius 10.5 cm and sector angle 60°?
Answer: r = 10.5 cm , \( \theta = 60^\circ \)
Perimeter of the sector = \( 2r + \frac{2\pi r \theta}{360^\circ} \)
\( = 10.5 \times 2 + 2 \times \frac{22}{7} \times \frac{10.5 \times 60}{360} \)
= 21 + 11 = 32 cm

Question. A sector of a circle of radius 8 cm contains an angle of 135. Find the area of the sector.
Answer: It is given that the radius of circle = 8 cm and angle, \( \theta = 135^\circ \)
Therefore, area of sector = \( \frac{\theta}{360^\circ} \times \pi r^2 \)
\( = \frac{135^\circ}{360^\circ} \times \pi \times 8 \times 8 \)
\( = \frac{135^\circ}{360^\circ} \times \pi \times 64 \)
\( = 24\pi \) cm\(^2\)

Question. The circumference of two circles are in the ratio 4 : 9. Find the ratio of their area.
Answer: \( \frac{2\pi r_1}{2\pi r_2} = \frac{4}{9} \)

\( \implies \frac{r_1}{r_2} = \frac{4}{9} \)
Now, \( \frac{\pi r_1^2}{\pi r_2^2} = \left( \frac{r_1}{r_2} \right)^2 = \left( \frac{4}{9} \right)^2 = \frac{16}{81} \)
Ratio of areas = 16 : 81

Question. The wheels of a car make 2500 revolutions in covering a distance of 4.95 km. Find the diameter of a wheel.
Answer: Distance covered by the wheel in 1 revolution = \( \left( \frac{4.95 \times 1000 \times 100}{2500} \right) \text{ cm} \)
= 198 cm
\( \therefore \) The circumference of the wheel = 198 cm
Let the diameter of the wheel be d cm
Then, \( \pi d = 198 \)

\( \implies \frac{22}{7} \times d = 198 \)

\( \implies d = \frac{198 \times 7}{22} \)
= 63 cm
Hence diameter of the wheel is 63 cm.

Question. The area of a sector of a circle of radius 5 cm is \( 5\pi \text{ cm}^2 \). Find the angle contained by the sector.
Answer: Radius of circle = 5cm
Area of sector = \( 5\pi \text{ cm}^2 \)
\( \therefore \) Area of sector = \( \frac{\theta}{360^\circ} \times \pi r^2 \)

\( \implies 5\pi = \frac{\theta}{360^\circ} \times \pi \times 5 \times 5 \)

\( \implies \theta = \frac{5\pi \times 360^\circ}{\pi \times 5 \times 5} = 72^\circ \)

Question. The area enclosed between the concentric circles is 770 cm\(^2\). If the radius of the outer circle is 21 cm, find the radius of the inner circle.
Answer: Radius of outer circle (R) = 21cm
let radius of inner circle = r cm
Given,
Area enclosed between concentric circle = 770cm\(^2\)

\( \implies \pi R^2 - \pi r^2 = 770 \)

\( \implies \frac{22}{7} \times 21 \times 21 - \frac{22}{7} \times r^2 = 770 \)

\( \implies \frac{22}{7} (441 - r^2) = 770 \)

\( \implies 441 - r^2 = \frac{770 \times 7}{22} \)

\( \implies 441 - r^2 = 245 \)

\( \implies r^2 = 441 - 245 \)

\( \implies r^2 = 196 \)

\( \implies r = \sqrt{196} = 14 \text{ cm} \)
\( \therefore \) radius of inner circle = 14cm

Question. Prove that the area of a circular path of uniform width h surrounding a circular region of radius r is \( \pi h(2r + h) \).
Answer: Given, radius of inner circle = r
width of path = h
Then radius of outer circle = R = r + h
\( \therefore \) Area of path = Area of outer circle - Area of inner circle
= \( \pi(R)^2 - \pi(r)^2 \)
= \( \pi(r + h)^2 - \pi r^2 \)
= \( \pi(r^2 + h^2 + 2rh) - \pi r^2 \)
= \( \pi r^2 + \pi h^2 + 2\pi rh - \pi r^2 \)
= \( \pi h^2 + 2\pi rh \)
= \( \pi h(h + 2r) \)

Question. The inner perimeter of a racing track is 400 m and the outer perimeter is 488 m. The length of each straight portion is 90 m and the end are semicircles. Find the cost of developing the track at the rate of Rs 12.50/m\(^2\).
Answer: Perimeter of 2 inner semicircles = (400 - 2 \( \times \) 90) m = (400 - 180) m = 220 m
Radius of each semicircle = \( \frac{220}{2\pi} = \frac{220 \times 7}{2 \times 22} = 35 \) m
Perimeter of 2 outer semicircles = (488 - 180) m = 308 m
Radius of each outer semicircle = \( \frac{308}{2\pi} = \frac{308 \times 7}{2 \times 22} = 49 \) m
Width of the track = outer radius - inner radius = 49 - 35 = 14 m
Area of rectangular tracks = 2 x area of rectangle
= 2 \( \times l \times b \) = 2 \( \times \) 90 \( \times \) 14 = 2520 m\(^2\)
Area of two semicircular rings = area of one circular ring
= \( \pi (R^2 - r^2) = \frac{22}{7} (49^2 - 35^2) \text{ m}^2 \)
= \( \frac{22}{7} \times (49 - 35)(49 + 35) \text{ m}^2 \)
= \( \frac{22}{7} \times 14 \times 84 = 3696 \text{ m}^2 \)
Total area of track = (2520 + 3696) m\(^2\) = 6216 m\(^2\)
Cost of developing the track at the rate of Rs 12.50/ m\(^2\) = Rs 6216 \( \times \) 12.50 = Rs 77,700.

Question. Find upto three places of decimal the radius of the circle whose area is the sum of the areas of two triangles whose sides are 35, 53, 66 and 33, 56, 65 measured in centimetres (Use \( \pi = 22/7 \)).
Answer: We have to find upto three places of decimal the radius of the circle whose area is the sum of the areas of two triangles whose sides are 35, 53, 66 and 33, 56, 65 measured in centimetres.
For the first triangle, we have a = 35, b = 53 and c = 66.
\( \therefore s = \frac{a+b+c}{2} = \frac{35+53+66}{2} = 77 \)cm
Let \( \Delta_1 \) be the area of the first triangle. Then,
\( \Delta_1 = \sqrt{s(s - a)(s - b)(s - c)} \)

\( \implies \Delta_1 = \sqrt{77(77 - 35)(77 - 53)(77 - 66)} = \sqrt{77 \times 42 \times 24 \times 11} \)

\( \implies \Delta_1 = \sqrt{7 \times 11 \times 7 \times 6 \times 6 \times 4 \times 11} \)
= \( \sqrt{7^2 \times 11^2 \times 6^2 \times 2^2} = 7 \times 11 \times 6 \times 2 = 924 \text{ cm}^2 \) — (i)
For the second triangle, we have a = 33, b = 56, c = 65.
\( \therefore s = \frac{a+b+c}{2} = \frac{33+56+65}{2} = 77 \)cm
Let \( \Delta_2 \) be the area of the second triangle. Then,
\( \Delta_2 = \sqrt{s(s - a)(s - b)(s - c)} \)

\( \implies \Delta_2 = \sqrt{77(77 - 33)(77 - 56)(77 - 65)} \)

\( \implies \Delta_2 = \sqrt{77 \times 44 \times 21 \times 12} \)
= \( \sqrt{7 \times 11 \times 4 \times 11 \times 3 \times 7 \times 3 \times 4} = \sqrt{7^2 \times 11^2 \times 4^2 \times 3^2} \)

\( \implies \Delta_2 = 7 \times 11 \times 4 \times 3 = 924 \text{ cm}^2 \)
Let r be the radius of the circle. Then,
Area of the circle = Sum of the areas of two triangles

\( \implies \pi r^2 = \Delta_1 + \Delta_2 \)

\( \implies \pi r^2 = 924 + 924 \)

\( \implies \frac{22}{7} \times r^2 = 1848 \)

\( \implies r^2 = 1848 \times \frac{7}{22} = 3 \times 4 \times 7 \times 7 \)
\( r = \sqrt{3 \times 2^2 \times 7^2} = 2 \times 7 \times \sqrt{3} = 14\sqrt{3} \) cm

Question. If the radius of the circle is \( \frac{7}{\sqrt{\pi}} \) cm, then its area is 
(a) 98sq. cm
(b) 45sq. cm
(c) 22 sq. cm
(d) 49 sq. cm
Answer: (d) 49 sq. cm
Explanation: Area of the circle = \( \pi r^2 \)

\( \implies \) Area of the circle = \( \pi \left( \frac{7}{\sqrt{\pi}} \right)^2 = \pi \times \frac{49}{\pi} = 49 \) sq. cm

Question. If the circumference of a circle and the perimeter of a square are equal, then 
(a) Area of the circle = \( \frac{1}{2} \) Area of the square
(b) area of the circle > area of the square
(c) area of the circle < area of the square
(d) area of the circle = area of the square
Answer: (b) area of the circle > area of the square
Explanation: Let the radius of the circle be r and side of the square be a.
Then, according to the question,
\( 2\pi r = 4a \)

\( \implies a = \frac{2\pi r}{4} = \frac{\pi r}{2} \)..........(i)
Now, the ratio of their areas,
\( \pi r^2 \) and \( a^2 \)

\( \implies \pi r^2 \) and \( \left( \frac{\pi r}{2} \right)^2 \) [From eq. (i)]

\( \implies \pi r^2 \) and \( \frac{\pi^2 r^2}{4} \)
Therefore, Area of the circle > Area of the square

Question. If a chord subtends an angle of 60° at the centre, then the area of the corresponding segment is 
(a) \( \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) r^2 \text{ sq. units} \)
(b) \( \left( \frac{\pi}{2} - \frac{\sqrt{3}}{2} \right) r^2 \text{ sq. units} \)
(c) \( \left( \frac{\pi}{6} + \frac{\sqrt{3}}{2} \right) r^2 \text{ sq. units} \)
(d) \( \left( \frac{\pi}{2} + \frac{\sqrt{3}}{2} \right) r^2 \text{ sq. units} \)
Answer: (a) \( \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) r^2 \text{ sq. units} \)
Explanation: Area of segment = \( \frac{\theta}{360^\circ} \times \pi r^2 - \frac{1}{2} r^2 \sin \theta \)

\( \implies \) Area of segment = \( \frac{60^\circ}{360^\circ} \times \pi r^2 - \frac{1}{2} r^2 \sin 60^\circ \)

\( \implies \) Area of segment = \( \frac{\pi r^2}{6} - \frac{r^2}{2} \times \frac{\sqrt{3}}{2} \)

\( \implies \) Area of segment = \( \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) r^2 \text{ sq. units} \)

Question. The length of the minute hand of a clock is 14cm. The area swept by the minute hand in 1 hour is 
(a) 616 sq. cm
(b) 516 sq. cm
(c) 628 sq. cm
(d) 542 sq. cm
Answer: (a) 616 sq. cm
Explanation: \( \because \) Angle describe by minute hand in 1 hour = 360°
\( \therefore \theta = 360^\circ \)
\( \therefore \) Area of the sector = \( \frac{\theta}{360^\circ} \times \pi r^2 = \frac{360^\circ}{360^\circ} \times \frac{22}{7} \times 14 \times 14 = 616 \) sq. cm

Question. A horse is tied to a peg at one corner of a square-shaped gross field of side 25 m by means of a 14m long rope. The area of that part of the field in which the horse can graze is 
(a) 156 sq. cm
(b) 142 sq. cm
(c) 102 sq. cm
(d) 128 sq. cm
Answer: (c) 102 sq. cm
Explanation: Area of the shaded region = \( \frac{\theta}{360} \times \pi r^2 \)

\( \implies \) Area of the shaded region = \( \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times 14 \times 14 \)

\( \implies \) Area of the shaded region = 102 sq. cm

Question. What is the perimeter of a sector of angle 45° of a circle with radius 7 cm? 
Answer: Given radius r of circle = 7 cm.
l = length of the arc = \( \frac{\theta \pi r}{180^\circ} \)
\( = \frac{45^\circ}{180^\circ} \times \frac{22}{7} \times 7 = \frac{11}{2} \)cm
Hence, perimeter of sector = (r + r+ l)
\( = (7 + 7 + \frac{11}{2}) \) cm
\( = (14 + \frac{11}{2}) \) cm = \( (\frac{28+11}{2}) \) cm
\( = \frac{39}{2} \) cm = 19.5 cm

Question. Radius of a circle is 1m. If diameter is increased by 100% then find the percentage increase in its area. 
Answer: Area of circle = \( \pi \text{ m}^2 \)
New diameter = 2 m + \( \frac{100}{100} \times 2 \) m = 4 m
New radius = 2 m
New area = \( 4\pi \text{ m}^2 \)
Increase in area = \( 4\pi - \pi = 3\pi \text{ m}^2 \)
% increase in area = \( \frac{3\pi}{\pi} \times 100 = 300\% \)

Question. If the perimeter of a semicircular protractor is 66 cm, find the radius of the protractor. 
Answer: Let radius of the protractor be r cm.
Perimeter = \( [\frac{1}{2} \times 2\pi r + 2r] \)
66 cm = \( [\pi r + 2r] \)

\( \implies 66 = (\pi + 2)r \)

\( \implies 66 = \left( \frac{22}{7} + 2 \right)r = \left( \frac{22+14}{7} \right)r \)

\( \implies 66 = \left( \frac{36}{7} \right)r \)

\( \implies \frac{7 \times 66}{36} = r \implies r = \frac{77}{6} \) cm

Question. If the circumference is numerically equal to 3 times the area of a circle, then find the radius of the circle. 
Answer: Let radius = r units
According to the question,
\( 2\pi r = 3\pi r^2 \)

\( \implies r = \frac{2}{3} \) units

Question. The area of two concentric circles forming a ring are 154 cm\(^2\) and 616 cm\(^2\). Find the breadth of the ring. 
Answer: Area of bigger circle = \( \pi R^2 = 616 \)

\( \implies R = 14 \) cm
and area of smaller circle = \( \pi r^2 = 154 \)

\( \implies r = 7 \) cm
Breadth of the ring = 14 - 7 = 7 cm

Question. The area of a sector of a circle of radius 2 cm is \( \pi \) cm\(^2\). Find the angle contained by the sector. 
Answer: Radius of circle = 2cm
Area of sector = \( \pi \text{ cm}^2 \)
\( \therefore \) Area of sector = \( \frac{\theta}{360^\circ} \times \pi r^2 \)

\( \implies \pi = \frac{\theta}{360^\circ} \times \pi \times 2 \times 2 \)

\( \implies \theta = \frac{\pi \times 360^\circ}{\pi \times 2 \times 2} = 90^\circ \)

Question. Find the area of a quadrant of a circle , whose circumference is 22 cm. 
Answer: Given , Circumference = 22 cm

\( \implies 2\pi r = 22 \)

\( \implies r = \frac{7}{2} = 3.5\text{cm} \)
Area of Circle = \( \pi r^2 = \frac{22}{7} \times (3.5)^2 = 38.5\text{cm}^2 \)
Area of quadrant of circle = \( \frac{\text{Area of circle}}{4} \)
= \( \frac{38.5}{4} = 9.625\text{cm}^2 \)
\( \therefore \) Area of the quadrant of cicle = 9.625 cm\(^2\)

Question. A semicircular region and a square region have equal perimeters. The area of the square region exceeds that of the semicircular region by 4 cm\(^2\). Find the perimeters and areas of the two regions. 
Answer: Let radius of semicircular region be r units.
Perimeter = 2r + \( \pi r \)
Let side of square be x units
Perimeter = 4x units.
A.T.Q, 4x = 2r + \( \pi r \)

\( \implies x = \frac{2r+\pi r}{4} \)
Area of semicircle = \( \frac{1}{2} \pi r^2 \)
Area of square = \( x^2 \)
A.T.Q, \( x^2 = \frac{1}{2} \pi r^2 + 4 \)

\( \implies \left( \frac{2r+\pi r}{4} \right)^2 = \frac{1}{2} \pi r^2 + 4 \)

\( \implies \frac{1}{16} (4r^2 + \pi^2 r^2 + 4\pi r^2) = \frac{1}{2} \pi r^2 + 4 \)

\( \implies 4r^2 + \pi^2 r^2 + 4\pi r^2 = 8\pi r^2 + 64 \)

\( \implies 4r^2 + \pi^2 r^2 - 4\pi r^2 = 64 \)

\( \implies r^2 (4 + \pi^2 - 4\pi) = 64 \)

\( \implies r^2 (\pi - 2)^2 = 64 \)

\( \implies r = \sqrt{\frac{64}{(\pi-2)^2}} \)

\( \implies r = \frac{8}{\pi-2} = \frac{8}{\frac{22}{7}-2} = 7 \) cm
Perimeter of semicircle = \( 2 \times 7 + \frac{22}{7} \times 7 = 36 \) cm
Perimeter of square = 36 cm
Side of square = \( \frac{36}{4} = 9 \) cm
Area of square = \( 9 \times 9 = 81 \text{ cm}^2 \)
Area of semicircle = \( \frac{\pi r^2}{2} = \frac{22}{2 \times 7} \times 7 \times 7 = 77 \text{ cm}^2 \)

Question. The angle described by the minute hand between 4.00 pm and 4.25 pm is 
(a) 90°
(b) 150°
(c) 125°
(d) 100°
Answer: (b) 150°
Explanation: Time duration between 4.00 pm and 4.25 pm = 25 minutes
Angle described by minute hand in 60 minutes = \( 360^\circ \)

\( \implies \) Angle described by minute hand in 25 minutes = \( \frac{360^\circ}{60^\circ} \times 25 = 150^\circ \)

Question. Four circles each of radius ‘a’ touch each other. The area between them is 
(a) None of the options
(b) \( \frac{6}{7} a^2 \)
(c) \( \frac{6}{7} a \)
(d) \( \frac{7}{6} a^2 \)
Answer: (b) \( \frac{6}{7} a^2 \)
Explanation: Area of required region = Area of square – Area of 4 quadrant

\( \implies \) Area of required region = \( (2a)^2 - 4 \times \frac{1}{4} \times \frac{22}{7} \times a^2 \)
= \( 4a^2 - \frac{22}{7} a^2 \)
= \( \frac{6}{7} a^2 \) sq. cm

Question. The area of a square that can be inscribed in a circle of radius 10 cm is 
(a) 100 sq. cm
(b) 300 sq. cm
(c) 200 sq. cm
(d) 150 sq. cm
Answer: (c) 200 sq. cm
Explanation: Given: Radius (r) = 10 cm
Let the side of the square be x cm
Now, using Pythagoras theorem,
\( x^2 + x^2 = (2r)^2 \)
\( 2x^2 = (20)^2 \)

\( \implies 2x^2 = 400 \)
\( x^2 = 200 \) sq. cm
Therefore, the area of the square = 200 sq. cm.

Question. The radius of the circle whose area is equal to the sum of the areas of the two circles of radii 24cm and 7cm is 
(a) 24cm
(b) 7cm
(c) 25cm
(d) 31cm
Answer: (c) 25cm
Explanation: Let required radius be R.
Then according to the question,
\( \pi R^2 = \pi r_1^2 + \pi r_2^2 \)
= \( \pi (r_1^2 + r_2^2) \)

\( \implies R^2 = r_1^2 + r_2^2 \)

\( \implies R^2 = (24)^2 + (7)^2 = 576 + 49 = 625 \)

\( \implies R = 25 \) cm

Question. Area of the largest circle that can be inscribed in a semicircle of radius ‘r’ units is 
(a) \( \frac{r^2}{\sqrt{2}} \) sq.units
(b) \( \frac{r^2}{2} \) sq.units
(c) \( (\pi/4) r^2 \) sq.units
(d) \( \sqrt{2}r^2 \) sq.units
Answer: (c) \( (\pi/4) r^2 \) sq.units
Explanation: Here, Diameter of circle = Radius of semicircle = r
\( \therefore \) Radius of the circle = \( \frac{r}{2} \)
\( \therefore \) Area of the circle = \( \pi \left( \frac{r}{2} \right)^2 = \frac{\pi r^2}{4} \)
\( \therefore (\pi/4) r^2 \) sq.units

Question. Find the ratio of the area of the incircle and circumcircle of a square. 
Answer: Let the side of square = x units
Diagonal of the square = \( \sqrt{2}x \) units
Diameter of the incircle = x units
Diameter of the circumcircle = \( \sqrt{2}x \) units
\( \frac{\text{Area of incircle}}{\text{Area of circumcircle}} = \frac{\pi \left( \frac{x}{2} \right)^2}{\pi \left( \frac{\sqrt{2}x}{2} \right)^2} = \frac{x^2/4}{2x^2/4} = \frac{1}{2} \)
Ratio = 1 : 2

Question. What will be the perimeter of a quadrant of a circle of radius r cm? 
Answer: Circumference of circle = \( 2\pi r \)
length of the arc of quadrant = \( \frac{1}{4} \times 2\pi r = \frac{\pi r}{2} \)
Perimeter of a quadrant = \( r + r + \frac{1}{4} \times 2\pi r = 2r + \frac{1}{2} \times \pi r = \frac{r}{2} (\pi + 4) \) cm
Therefore, perimeter of the quadrant with radius "r" = \( \frac{r}{2} (\pi + 4) \) cm

Question. What is the diameter of a circle whose area is equal to the sum of the areas of two circles of radii 40 cm and 9 cm? 
Answer: Area of the circle = sum of areas of two circles
\( \pi r^2 = \pi \times (40)^2 + \pi(9)^2 \)
or, \( r^2 = 1600 + 81 \)
or, \( r = \sqrt{1681} \)
= 41cm.
Diameter of circle is double of radius.
\( \therefore \) Diameter of given circle = \( 2 \times r \)
= 41 \(\times\) 2
= 82 cm.

Question. The area of the sector of a circle of radius 10.5 cm is 69.3 cm\(^2\). Find the central angle of the sector. 
Answer: It is given that area of the sector = 69.3 cm\(^2\)
and Radius = 10.5 cm
Now, Area of the sector = \( \frac{\pi r^2 \theta}{360} \)

\( \implies \frac{\pi \times (10.5)^2 \times \theta}{360} = 69.3 \)

\( \implies \theta = \frac{69.3 \times 360 \times 7}{10.5 \times 10.5 \times 22} = 72^\circ \)
Therefore, Central angle of the sector = 72°

Question. A garden roller has a circumference of 4 m. Find the number of revolutions, it makes in moving 40 m. 
Answer: Given, Circumference of circle = 4 m
Total distance covered by roller = 40 m
Distance covered in one Revolution = circumference of roller = 4m
Number of revolution = \( \frac{\text{Total Distance covered by roller}}{\text{Distance covered in one revolution}} \)
= 40/4 = 10 revolutions
Total revolution = 10.

Question. A sector is cut from a circle of radius 21 cm. The angle of the sector is 150°. Find the length of its arc and area. 
Answer: The arc length l and area A of a sector of angle \( \theta \) in a circle of radius r are is given by
\( l = \frac{\theta}{360} \times 2\pi r \) and \( A = \frac{\theta}{360} \times \pi r^2 \) respectively.
Here, r = 21 cm and \( \theta \) = 150
\( \therefore l = \left\{ \frac{150}{360} \times 2 \times \frac{22}{7} \times 21 \right\} \text{ cm} = 55\text{cm} \)
and, \( A = \left\{ \frac{150}{360} \times \frac{22}{7} \times (21)^2 \right\} \text{ cm}^2 = \frac{1155}{2} \text{cm}^2 = 577.5 \text{ cm}^2 \)

Question. A circular grassy plot of land, 42 m in diameter, has a path 3.5 m wide running around it on the outside. Find the cost of travelling the path at Rs.4 per square metre. 
Answer: Radius of the plot = 21 m. Radius of the plot including the path = (21 + 3.5) m = 24.5 m
\( \therefore \) Area of the path
= \( \{ \pi(24.5)^2 - \pi(21)^2 \} \text{ m}^2 \)
= \( \pi \{ (24.5 + 21) (24.5 - 21) \} \text{ m}^2 \)
= \( \{ \pi (45.5) \times (3.5) \} \text{ m}^2 \)
= \( \frac{22}{7} \times 45.5 \times 3.5 \text{m}^2 = 500.5 \text{ m}^2 \)
Hence, cost of gravelling the path = Rs (500.5 \(\times\) 4 ) = Rs 2002.

Question. The outer circumference of a circular race-track is 528 m. The track is everywhere 14 m wide. Calculate the cost of levelling the track at the rate of 50 paise per square metre (Use \( \pi = \frac{22}{7} \)) 
Answer: Let, radius of inner circle = rm
radius of outer circle = Rm
width of track = 14m
Given,
Outer circumference = 528m

\( \implies 2\pi R = 528m \)

\( \implies 2 \times \frac{22}{7} \times R = 528 \)

\( \implies R = \frac{528 \times 7}{2 \times 22} = 84m \)
the inner radius = 84 - 14 = 70m
\( \therefore \) Area of track = Area of outer circle - Area of inner circle
= \( \frac{22}{7} \times 84 \times 84 - \frac{22}{7} \times 70 \times 70 \)
= \( \frac{22}{7} \times 7 \times 7 [12 \times 12 - 10 \times 10] \)
= 6776 m\(^2\)
Rate of levelling the track = 50 paise per m\(^2\)
\( \therefore \) Total cost of levelling = 6776 \(\times\) 50
= 338800 paise
= Rs 3388

Question. AB is one of the direct common tangent of two circles of radii 12 cm and 4 cm respectively touching each other. Find the area of the region enclosed by the circles and the tangent. 
Answer: Let tangent touches the circles with centre O and O' at the points A and B respectively.

\( \implies \angle 1 = \angle 2 = 90^\circ \) [tangent makes right angle with the radius at the point of contact]
Through O' draw \( O'M || AB \) meeting OA in M
Thus, ABO'M is a rectangle with AM = O'B = 4 cm
In triangle OMO',
OM = OA - AM = (12 - 4) cm = 8 cm
and OO' = OP + O'P = 12 + 4 = 16 cm
\( OO'^2 = OM^2 + (O'M)^2 \)
\( x^2 = 16^2 - 8^2 = (16 - 8)(16 + 8) \)
= 8 \(\times\) 24 = 8 \(\times\) 8 \(\times\) 3

\( \implies x = \sqrt{8 \times 8 \times 3} = 8\sqrt{3} \) cm
Area ABO'O = area of rectangle ABO'M + area of \( \triangle OMO' \)
= \( AM \times O'M + \frac{1}{2} \times OM \times O'M \)
= \( 4 \times 8\sqrt{3} + \frac{1}{2} \times 8 \times 8\sqrt{3} \)
= \( 32\sqrt{3} + 32\sqrt{3} = 64\sqrt{3} \) cm
Now, \( \triangle OMO' \) is a right triangle with \( \angle M = 90^\circ \)
\( \frac{O'M}{OO'} = \frac{8\sqrt{3}}{16} = \cos(\angle 3) \)

\( \implies \angle 3 = 30^\circ \) and \( \frac{OM}{OO'} = \cos \angle 4 \)

\( \implies \frac{8}{16} = \cos \angle 4 \)
\( \angle 4 = 60^\circ \)

\( \implies \angle AOP = 60^\circ \)
= angle of sector AOP
\( \angle 3 = 30^\circ \implies \angle MO'O = 30^\circ \)

\( \implies \angle BO'P = 90^\circ + \angle MO'O = 90^\circ + 30^\circ \)
= 120° = angle of sector BOP
Area of the portion enclosed between the circles and the tangents is
= area of trapezium ABO'O - 2(area of the sectors)
= \( 64\sqrt{3} \text{ cm}^2 \) - (area of sector AOP + area of sector BO'P)
= \( 64\sqrt{3}\text{cm}^2 - \left[ \frac{60^\circ}{360^\circ} \times \pi \times 12^2 + \frac{120^\circ}{360^\circ} \times \pi \times 4^2 \right] \text{ cm}^2 \)
= \( 64\sqrt{3}\text{cm}^2 - \frac{\pi}{6} [144 + 2 \times 16]\text{cm}^2 \)
= \( 64\sqrt{3}\text{cm}^2 - \frac{\pi}{6} \times 8[18 + 2 \times 2]\text{cm}^2 \)
= \( 64\sqrt{3}\text{cm}^2 - \frac{4\pi}{3} \times 2[9 + 2]\text{cm}^2 \)
= \( \left[ 64\sqrt{3} - \frac{8}{3} \times \pi \times 11 \right] \text{ cm}^2 \)
= \( [64 \times 1.73 - \frac{88}{3} \times \frac{22}{7}] \text{ cm}^2 \)
[110.72 - 92.1904] = 18.5296 cm\(^2\)

Question. Find the difference of the areas of two segments of a circle formed by a chord of length 5 cm subtending angle of 90° at the centre. 
Answer: Chord AB = 5 cm divides the circle into two segments minor segment APB and major segment AQB. We have to find out the difference in area of major and minor segment.
Here, we are given that \( \theta = 90^\circ \)
Area of \( \triangle OAB = \frac{1}{2} \text{ Base} \times \text{Altitude} = \frac{1}{2} r \times r = \frac{1}{2}r^2 \)
Area of minor segment APB
= \( \frac{\pi r^2 \theta}{360^\circ} - \text{Area of } \triangle OAB \)
= \( \frac{\pi r^2 90^\circ}{360^\circ} - \frac{1}{2}r^2 \)

\( \implies \) Area of minor segment = \( \left( \frac{\pi r^2}{4} - \frac{r^2}{2} \right) \) …(i)
Area of major segment AQB = Area of circle – Area of minor segment
= \( \pi r^2 - \left[ \frac{\pi r^2}{4} - \frac{r^2}{2} \right] \)

\( \implies \) Area of major segment AQB = \( \left[ \frac{3}{4}\pi r^2 + \frac{r^2}{2} \right] \) …(ii)
Difference between areas of major and minor segment
= \( \left( \frac{3}{4}\pi r^2 + \frac{r^2}{2} \right) - \left( \frac{\pi r^2}{4} - \frac{r^2}{2} \right) \)
= \( \frac{3}{4}\pi r^2 + \frac{r^2}{2} - \frac{\pi r^2}{4} + \frac{r^2}{2} \)

\( \implies \) Required area = \( \frac{2}{4}\pi r^2 + r^2 = \frac{1}{2}\pi r^2 + r^2 \)
In right \( \triangle OAB \),
\( r^2 + r^2 = AB^2 \)

\( \implies 2r^2 = 5^2 \)

\( \implies r^2 = \frac{25}{2} \)
Therefore, required area = \( \left[ \frac{1}{2} \pi \times \frac{25}{2} + \frac{25}{2} \right] = \left[ \frac{25}{4}\pi + \frac{25}{2} \right] \) cm\(^2\)