CBSE Class 10 Mathematics Area Related To Circle Worksheet Set 04

Question. Find the circumference and area of a circle of radius 4.2 cm.
Answer: 26.4 cm, 55.44 \( cm^{2} \)

Question. Find the circumference of a circle whose area is 301.84 \( cm^{2} \).
Answer: 61.6 cm

Question. Find the area of a circle whose circumference is 44 cm.
Answer: 154 \( cm^{2} \)

Question. The circumference of a circle exceeds the diameter by 16.8 cm. Find the circumference of the circle.
Answer: 24.64 cm

Question. A horse is tied to a pole with 28 m long string. Find the area where the horse can graze. (Take \( \pi = 22/7 \)).
Answer: 2464 \( m^{2} \)

Question. A steel wire when bent in the form of a square encloses an area of 121 \( cm^{2} \). If the same wire is bent in the form of a circle, find the area of the circle.
Answer: 154 \( cm^{2} \)

Question. A sector of a circle of radius 4 cm contains an angle of 30º. Find the area of the sector.
Answer: \( cm^{2} \)

Question. A sector of a circle of radius 8 cm contains an angle of 135º. Find the area of the sector.
Answer: 24 \( \pi cm^{2} \)

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: 90°

Question. A play ground has the shape of a rectangle, with two semi-circles on its smaller sides as diameters, added to its outside. If the sides of the rectangle are 36 m and 24.5 m, find the area of the playground. (Take \( \pi = 22/7 \)).
Answer: 1353.625 \( m^{2} \)

Question. The outer circumference of a circular race track is 528 m. The track is everywhere 14m wide. Calculate the cost of levelling the track at the rate of 50 paise per square metre (Use \( \pi = 22/7 \)).
Answer: Rs. 3388

Question. A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants of radii 3.5 m have been cut. Find the area of the remaining part.
Answer: 261.5 \( m^{2} \)

Question. The circumference of two circles are in the ratio 2 : 3. Find the ratio of their areas.
Answer: 4 : 9

Question. The side of a square is 10 cm. Find the area of circumscribed and inscribed circles.
Answer: 157 \( m^{2} \), 78.5 \( m^{2} \)

Question. The sum of the radii of two circles is 140 cm and the difference of their circumferences is 88 cm. Find the diameters of the circles.
Answer: 154 cm, 126 cm

Question. The area of a circle inscribed in an equilateral triangle is 154 \( cm^{2} \). Find the perimeter of the triangle. [Use \( \pi = 22/7 \) and \( \sqrt{3} = 1.73 \)]
Answer: 72.7 cm

Question. A field is in the form of a circle. A fence is to be erected around the field. The cost of fencing would be Rs. 2640 at the rate of Rs. 12 per metre. Then, the field is to be thoroughly ploughed at the cost of Rs. 0.50 per \( m^{2} \). What is the amount required to plough the field ? [Take \( \pi = 22/7 \)].
Answer: Rs. 1925

Question. If a square is inscribed in a circle, find the ratio of the areas of the circle and the square.
Answer: \( \pi \) : 2

Question. A park is in the form of a rectangle 120m × 100 m. At the centre of the park there is a circular lawn. The area of park excluding lawn is 8700 \( m^{2} \). Find the radius of the circular lawn. (Use \( \pi = 22/7 \))
Answer: 21.41m

Question. The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having its area equal to the sum of the areas of the two circles.
Answer: 10 cm

Question. The radii of two circles are 19 cm and 9 cm respectively. Find the radius and area of the circle which has its circumference equal to the sum of the circumferences of the two circles.
Answer: 28 cm, 2464 \( cm^{2} \)

Question. A car travels 1 kilometre distance in which each wheel makes 450 complete revolutions. Find the radius of the its wheels.
Answer: 35.35 cm

Question. The area of enclosed between two concentric circles is 770 \( cm^{2} \). If the radius of the outer circle is 21 cm, find the radius of the inner circle.
Answer: 14 cm

Question. Find, in terms of \( \pi \), the length of the arc that subtends an angle of 30º at the centre of a circle of radius 4 cm.
Answer: cm

Question. Find the angle subtended at the centre of a circle of radius 5 cm by an arc of length \( (5\pi/3) \) cm.
Answer: 60°

Question. An arc of length \( 20\pi \) cm subtends an angle of 144º at the centre of a circle. Find in terms of \( \pi \), the radius of the circle.
Answer: 25 cm

Question. An arc of length 15 cm subtends an angle of 45º at the centre of a circle. Find the radius of the circle.
Answer: cm

Question. Find the angle subtended at the centre of a circle of radius 'a' by an arc of length \( (a\pi/4) \) cm.
Answer: 45°

Question. A chord AB of a circle of radius 15 cm makes an angle of 60º at the centre of the circle. Find the area of the major and minor segment. (Take \( \pi = 3.14, \sqrt{3} = 1.73 \))
Answer: 20.295 \( cm^{2} \), 686.205 \( cm^{2} \)

Question. In a circle of radius 21 cm, an arc subtends an angle of 60º at the centre. Find:
(i) length of the arc
(ii) area of the sector formed by the arc
(iii) area of the segment formed by the corresponding chord of the arc.
Answer: (i) 22cm (ii) 231 \( cm^{2} \) (iii) 40.05 \( cm^{2} \)

Question. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find:
(i) area of the minor sector
(ii) area of the minor segment
(iii) area of the major sector
(iv) area of the major segment (Use \( \pi = 3.14 \))
Answer: (i) 78.5\( cm^{2} \) (ii) 28.5\( cm^{2} \) (iii) 235.5\( cm^{2} \) (iv) 285.5 \( cm^{2} \)

Question. A circular grassy plot of land, 42 m in diameter, has a path 3.5 m wide running round it on the outside. Find the cost of gravelling the path at Rs. 4 per square metre.
Answer: 500.5 \( m^{2} \)

Question. Find to the three places of decimals 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.
Answer: 14 cm

Question. In an equilateral triangle of side 24 cm, a circle is inscribed touching its sides. Find the area of the remaining portion of the triangle (Take \( \sqrt{3} = 1.732 \)).
Answer: 98.55 \( cm^{2} \)

Question. The part of the circular region enclosed by a chord and the corresponding arc of a circle is called 
(a) a segment
(b) a diameter
(c) a radius
(d) a sector
Answer: (a) a segment
Explanation: The part of the circular region enclosed by a chord and the corresponding arc of a circle is called a segment.

Question. If a line meets the circle in two distinct points, it is called 
(a) a chord
(b) a radius
(c) secant
(d) a tangent
Answer: (c) secant
Explanation: A secant line, also simply called a secant, is a line meet two points in a circle.

Question. Area of a sector of angle p (in degrees) of a circle with radius R is 
(a) \( \frac{p}{360} \times 2\pi R \)
(b) \( \frac{p}{180} \times \pi R^2 \)
(c) \( \frac{p}{180} \times 2\pi R \)
(d) \( \frac{p}{720} \times 2\pi R^2 \)
Answer: (d) \( \frac{p}{720} \times 2\pi R^2 \)
Explanation: Area of the sector of angle p of a circle with radius R
\( = \frac{\theta}{360} \times \pi r^2 = \frac{p}{360} \times \pi R^2 \)
\( = \frac{p}{2(360)} \times 2\pi R^2 = \frac{p}{720} \times 2\pi R^2 \)

Question. If ‘r’ is the radius of a circle, then it's circumference is given by 
(a) \( 2\pi r \)
(b) None of the options
(c) \( \pi r \)
(d) \( 2\pi d \)
Answer: (a) \( 2\pi r \)
Explanation: If the radius of a circle is given, the circumference or perimeter can be calculated using the formula below:-
Circumference = \( 2\pi r \)

Question. The perimeter of a protractor is 
(a) \( \pi r \)
(b) \( \pi r + 2r \)
(c) \( \pi + r \)
(d) \( \pi + 2r \)
Answer: (b) \( \pi r + 2r \)
Explanation: Let radius of the protractor be r.
\( \therefore \) Perimeter of protractor = Perimeter of semicircle + Diameter of semicircle

\( \implies \) Perimeter of protractor = \( \pi r + 2r \)

Question. If circumference of a circle is 44 cm, then what will be the area of the circle? 
Answer: Circumference of a circle = 44 cm
\( 2\pi r = 44 \)
\( 2 \times \frac{22}{7} \times r = 44 \)
\( \frac{44}{7} \times r = 44 \)
Radius of the circle = \( \frac{44 \times 7}{44} = 7 \)cm
Area of the circle = \( \pi r^2 = \frac{22}{7} \times 7 \times 7 \)
= 154 cm\(^2\).
So, Area of the circle is 154 cm\(^2\).

Question. Find the area of circle that can be inscribed in a square of side 10 cm. 
Answer: Side of square = 10 cm
Side of square = diameter of circle = 10 cm
Radius of the circle = \( \frac{10}{2} = 5 \)cm
Area of the circle = \( \pi \times r^2 \)
\( = \pi \times (5)^2 \)
\( = \pi \times 5 \times 5 \)
\( = 25\pi \) cm\(^2\)

Question. If the perimeter of a semi-circular protractor is 36 cm, then find its diameter. 
Answer: Perimeter of a semi-circular protactor = Perimeter of a semi-circle = \( \frac{1}{2} \)(circumference of circle) + diameter = \( \frac{1}{2} \)(circumference of circle) + 2 \( \times \) radius = \( (2r + \pi r) \)cm

\( \implies 2r + \pi r = 36 \) [Given, perimeter of semi-circular protactor = 36]

\( \implies r = \frac{36}{2 + \pi} \)

\( \implies r = 7 \)cm
Hence, diameter of semi-circular protactor = 2r = 2(7) = 14cm

Question. What is the perimeter of a square which circumscribes a circle of radius a cm? 
Answer: When a square circumscribes a circle, the radius of the circle is half the length of the square.
Therefore, if the radius of the circumscribed circle is a, the diameter will be 2a. It is this diameter that is equal to the length of the square.
Therefore, the length of the square is 2a cm.
Then area of a square = 4 \( \times \) side
= 4 \( \times \) 2a cm
= 8a cm

Question. On a square cardboard sheet of area 784 cm\(^2\), four circular plates of maximum size are placed such that each circular plate touches the other two plates and each side of the square sheet is tangent to circular plates. Find the area of the square sheet not covered by the circular plates.
Answer: Let the radius of each circular plate be r cm. Then,
Length of each side of the square sheet = 4r cm.
\( \therefore \) Area of the square cardboard sheet = \( (4r \times 4r) \) cm\(^2 = 16 r^2 \) cm\(^2\)
But, the area of the cardboard sheet is given to be 784 cm\(^2\)
\( 16r^2 = 784 \)

\( \implies r^2 = 49 \)

\( \implies r = 7 \)
Area of one circular plate = \( \pi r^2 = \frac{22}{7} \times 7^2 \) cm\(^2 = 154 \) cm\(^2\)
\( \therefore \) Area of four circular plates = \( 4 \times 154 \) cm\(^2 = 616 \) cm\(^2\)
\( \therefore \) Uncovered area of the square sheet = (784 - 616) cm\(^2 = 168 \) cm\(^2\)

Question. The circumference of a circle is 22 cm. Find the area of its quadrant. 
Answer: Suppose r be the radius of a circle
Circumference of a circle = 22cm

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

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

\( \implies r = \frac{7}{2} \)cm
Area of the quadrant of a circle = \( \frac{1}{4} \times \pi \times r^2 \)
\( = \left( \frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \right) \) cm\(^2 \)
\( = \frac{77}{8} \) cm\(^2 \)

Question. A sector of a circle of radius 4 cm contains an angle of 30°. Find the area of the sector. 
Answer: Radius of circle = 4cm
\( \theta = 30^\circ \)
\( \therefore \) Area of sector = \( \frac{\theta}{360^\circ} \times \pi r^2 \)
\( = \frac{30^\circ}{360^\circ} \times \pi \times 4 \times 4 \)
\( = \frac{4\pi}{3} \) cm\(^2 \)

Question. A circular pond is 17.5 m in diameter. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of Rs 25 per m\(^2\). 
Answer: \( \therefore \) Radius of a pond = \( \frac{17.5}{2} = 8.75 \)
\( \therefore \) Area of a pond = \( \pi(8.75)^2 \) sq. m
Radius of a circle including path = 8.75 + 2 = 10.75 m
According to question,
Area of the path = Area of a circle including path - Area of a pond
\( = \pi(10.75)^2 - \pi(8.75)^2 \)
\( = \pi [ (10.75)^2 - (8.75)^2 ] \)
\( = \frac{22}{7} [(10.75 + 8.75)(10.75 - 8.75)] \)
\( = \frac{22}{7} [19.5 \times 2] \)
\( = \frac{22}{7} \times 39 \)
\( = \frac{858}{7} \) sq. m
= 122.5 sq. m
Cost of constructing the path = \( 25 \times 122.5 = \) Rs. 3062.50

Question. A chord of a circle of radius 10cm subtends a right angle at the center. Find the area of the corresponding: (Use \( \pi = 3.14 \)) i. minor sector ii. major sector iii. minor segment iv. major segment 
Answer: i. Area of minor sector = \( \frac{\theta}{360} \pi r^2 \)
\( = \frac{90}{360} (3.14)(10)^2 \)
\( = \frac{1}{4} \times 3.14 \times 100 \)
\( = \frac{314}{4} \)
= 78.50 = 78.5 cm\(^2\)
ii. Area of major sector = Area of circle - Area of minor sector
= \( \pi(10)^2 - \frac{90}{360} \pi(10)^2 = 3.14 (100) - \frac{1}{4} (3.14)(100) \)
= 314 - 78.50 = 235.5 cm\(^2\)
iii. We know that area of minor segment
= Area of minor sector OAB - Area of \( \triangle OAB \)
\( \therefore \) area of \( \triangle OAB = \frac{1}{2} (OA)(OB) \sin \angle AOB \)
\( = \frac{1}{2} (OA)(OB) \) (\( \therefore \angle AOB = 90^\circ \))
Area of sector = \( \frac{\theta}{360} \pi r^2 \)
\( = \frac{1}{4} (3.14) (100) - 50 = 25(3.14) - 50 = 78.50 - 50 = 28.5 \) cm\(^2\)
iv. Area of major segment = Area of the circle - Area of minor segment
= \( \pi(10)^2 - 28.5 \)
= 100(3.14) - 28.5
= 314 - 28.5 = 285.5 cm\(^2\)

Please click on below link to download CBSE Class 10 Mathematics Area Related To Circle Worksheet Set D