CBSE Class 10 Mathematics Areas related to circles MCQs Set 08

Question. The area of a circle, whose circumference is 22 cm, is:
(a) \( 54 \text{ cm}^2 \)
(b) \( 46 \text{ cm}^2 \)
(c) \( 40.5 \text{ cm}^2 \)
(d) \( 38.5 \text{ cm}^2 \)
Answer: (d) \( 38.5 \text{ cm}^2 \)
Answer: Let \( r \) is the radius of the circle. Then,
\( 2\pi r = 22 \text{ cm} \)
\( \implies r = 22 / ( 2 \times 22/7) \)
\( \implies r = \frac{7}{2} \text{ cm} \) or \( 3.5 \text{ cm} \)
Thus, \( \pi r^2 = \frac{22}{7} \times 3.5 \times 3.5 \text{ cm}^2 = 38.5 \text{ cm}^2 \)

Question. If the sum of the areas of two circles with radii \( R_1 \) and \( R_2 \) is equal to the area of a circle of radius \( R \), then:
(a) \( R_1 + R_2 = R \)
(b) \( R_1^2 + R_2^2 = R^2 \)
(c) \( R_1 + R_2 < R \)
(d) \( R_1^2 + R_2^2 < R^2 \)
Answer: (b) \( R_1^2 + R_2^2 = R^2 \)
Answer: Explanation: According to the given condition,
Area of circle with radius \( R \) = Area of circle with radius \( R_1 \) + Area of circle with radius \( R_2 \)
\( \implies \pi R^2 = \pi R_1^2 + \pi R_2^2 \)
\( \implies R^2 = R_1^2 + R_2^2 \)

Question. If the sum of the circumferences of two circles with radii \( R_1 \) and \( R_2 \) is equal to the circumference of a circle of radius \( R \), then:
(a) \( R_1 + R_2 = R \)
(b) \( R_1 + R_2 > R \)
(c) \( R_1 + R_2 < R \)
(d) Nothing definite can be said about the relation among \( R_1 \), \( R_2 \) and \( R \).
Answer: (a) \( R_1 + R_2 = R \)
Answer: Explanation: According to the given condition,
Circumference of circle with radius \( R \) = Circumference of circle with radius \( R_1 \) + Circumference of circle with radius \( R_2 \)
\( \implies 2\pi R = 2\pi R_1 + 2\pi R_2 \)
\( \implies R = R_1 + R_2 \)

Question. It is proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be:
(a) 10 m
(b) 15 m
(c) 20 m
(d) 24 m
Answer: (a) 10 m
Answer: Explanation: Let the radius of the new park be \( R \).
\( \therefore \) Area of new park = Area of park I + Area of park II
Also, if \( r_1 \) and \( r_2 \) are the radius of circle
\( r_1 = \frac{d_1}{2} = \frac{16}{2} = 8 \)
\( r_2 = \frac{d_2}{2} = \frac{12}{2} = 6 \)
Area of the first circular park with diameter 16 m \( = \pi \left(\frac{16}{2}\right)^2 = \pi(8)^2 = 64\pi \text{ m}^2 \)
Area of the second circular park with diameter 12 m \( = \pi \left(\frac{12}{2}\right)^2 = \pi(6)^2 = 36\pi \text{ m}^2 \)
According to the given condition.
\( \pi R^2 = 64\pi + 36\pi \)

\( \implies \pi R^2 = 100\pi \)

\( \implies R^2 = 100 \)

\( \implies R = 10 \text{ m} \)

Question. The radii of two concentric circles are 4 cm and 5 cm. The difference in the areas of these two circles is:
(a) \( \pi \)
(b) \( 7\pi \)
(c) \( 9\pi \)
(d) \( 13\pi \)
Answer: (c) \( 9\pi \)
Answer: Required difference \( = \pi(5^2 - 4^2) = 9\pi \)

Question. If the area of a circle is \( 154 \text{ cm}^2 \), then its circumference is:
(a) 11 cm
(b) 22 cm
(c) 44 cm
(d) 55 cm
Answer: (c) 44 cm
Answer: Here, \( \pi r^2 = 154 \)

\( \implies r^2 = 154 \times 7/22 = 49 \)

\( \implies r = 7 \text{ cm} \)
So, perimeter \( = 2\pi r = 2 \times \frac{22}{7} \times 7 = 44 \text{ cm} \)

Question. A wire is in the shape of a circle of radius 21 cm. It is bent to form a square. The side of the square is: \( \left(\pi = \frac{22}{7}\right) \)
(a) 22 cm
(b) 33 cm
(c) 44 cm
(d) 66 cm
Answer: (b) 33 cm
Answer: Circumference of circle = Perimeter of the square.
So, \( 2\pi r = 4a \)

\( \implies 4a = 2 \times \frac{22}{7} \times 21 \)

\( \implies 4a = 132 \)

\( \implies a = 33 \text{ cm} \)

Question. The area of a circle that can be inscribed in a square of side 6 cm is:
(a) \( 36\pi \text{ cm}^2 \)
(b) \( 18\pi \text{ cm}^2 \)
(c) \( 12\pi \text{ cm}^2 \)
(d) \( 9\pi \text{ cm}^2 \)
Answer: (d) \( 9\pi \text{ cm}^2 \)
Answer: Explanation: It is given that the side of square = 6 cm
\( \implies \) Diameter of the circle inscribed in a square, \( d \) = Side of square = 6 cm
\( \therefore \) Radius of circle \( (r) = \frac{6}{2} = 3 \text{ cm} \)
\( \implies \) Area of circle \( = \pi r^2 = \pi(3)^2 = 9\pi \text{ cm}^2 \)

Question. The outer and inner diameters of a circular ring are 34 cm and 32 cm respectively. The area of the ring is:
(a) \( 66\pi \text{ cm}^2 \)
(b) \( 60\pi \text{ cm}^2 \)
(c) \( 33\pi \text{ cm}^2 \)
(d) \( 29\pi \text{ cm}^2 \)
Answer: (c) \( 33\pi \text{ cm}^2 \)
Answer: Area of the circular ring \( = \pi [17^2 - 16^2] \text{ cm}^2 = 33\pi \text{ cm}^2 \)

Question. The diameter of a circle whose area is equal to the sum of the areas of the two circles of radii 24 cm and 7 cm is:
(a) 31 cm
(b) 25 cm
(c) 62 cm
(d) 50 cm
Answer: (d) 50 cm
Answer: Explanation: Let, Radius of \( 1^{st} \) circle \( r_1 = 24 \text{ cm} \)
Area of \( 1^{st} \) circle \( = \pi(r_1)^2 = \pi(24)^2 = 576\pi \text{ cm}^2 \)
Radius of \( 2^{nd} \) circle \( r_2 = 7 \text{ cm} \)
Area of \( 2^{nd} \) circle \( = \pi(r_2)^2 = \pi(7)^2 = 49\pi \text{ cm}^2 \)
Let \( R \) be the radius of the new circle.
According to the given condition, Area of new circle = Area of \( 1^{st} \) circle + Area of \( 2^{nd} \) circle

\( \implies \pi R^2 = 576\pi + 49\pi \)

\( \implies R^2 = 625 \)

\( \implies R = 25 \text{ cm} \)
\( \therefore \) Diameter of new circle \( = 2R = 2 \times 25 = 50 \text{ cm} \)

Question. If a circular grass lawn of 35 m in radius has a path 7 m wide running around it on the outside, then the area of the path is:
(a) \( 1450 \text{ m}^2 \)
(b) \( 1576 \text{ m}^2 \)
(c) \( 1694 \text{ m}^2 \)
(d) \( 3368 \text{ m}^2 \)
Answer: (c) \( 1694 \text{ m}^2 \)
Answer: Explanation: Radius of outer concentric circle \( = (35 + 7) \text{ m} = 42 \text{ m} \)
Area of path \( = \pi(R^2 - r^2) = \frac{22}{7} \times (1764 - 1225) = \frac{22}{7} \times 539 = 1694 \text{ m}^2 \)

Question. The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36 cm and 20 cm is:
(a) 56 cm
(b) 42 cm
(c) 28 cm
(d) 16 cm
Answer: (c) 28 cm
Answer: Explanation: According to the given condition, Circumference of circle = Sum of circumference of two circle. Let \( r_1 \) and \( r_2 \) be the radius of the two circles then \( r_1 = \frac{36}{2} = 18 \text{ cm} \), \( r_2 = \frac{20}{2} = 10 \text{ cm} \).

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

\( \implies R = r_1 + r_2 = 18 + 10 = 28 \text{ cm} \)

Fill in the Blanks

Question. The ratio of the areas of a circle and an equilateral triangle whose diameter and a side are respectively equal is ......................
Answer: \( \frac{\pi}{\sqrt{3}} \)
Explanation: Given \( 2r = a \implies \frac{r}{a} = \frac{1}{2} \)
\( \frac{\text{Area of circle}}{\text{Area of equilateral } \Delta} = \frac{\pi r^2}{\frac{\sqrt{3}}{4}a^2} = \frac{4\pi}{\sqrt{3}} \times \frac{1}{4} = \frac{\pi}{\sqrt{3}} \)

Question. The radius of a wheel is 0.25 m. The number of revolutions it will make to travel a distance of 11 km is ......................
Answer: 7000
Explanation: Circumference of wheel \( = 2\pi r = 2 \times \frac{22}{7} \times 0.25 \)
Number of revolutions \( = \frac{\text{Distance Travelled}}{\text{Circumference of wheel}} = \frac{11 \times 1000}{2 \times \frac{22}{7} \times 0.25} = 7000 \)

Question. The area of the circle inscribed in a square of side a cm is ...................... .
Answer: \( \frac{\pi a^2}{4} \text{ cm}^2 \)
Explanation: Diameter of the circle = a

\( \implies \) Radius \( = \frac{a}{2} \)

\( \implies \) Area \( = \pi \left(\frac{a}{2}\right)^2 = \frac{\pi a^2}{4} \text{ cm}^2 \)

Question. If circumference and the area of a circle are numerically equal, then the diameter of the circle is ...................... .
Answer: 4 units
Explanation: Given, \( 2\pi r = \pi r^2 \)

\( \implies 2r = r^2 \implies r = 2 \)
Diameter \( = 2r = 2 \times 2 = 4 \text{ units.} \)

Question. If the circumference of a circle is 66 cm, then is its area is ...................... .
Answer: \( 86.625 \text{ cm}^2 \)
Explanation: Circumference = 66 cm
Let radius = \( r \)
\( 2\pi r = 66 \)

\( \implies r = \frac{66}{2\pi} = \frac{66 \times 7}{2 \times 22} = \frac{21}{4} \)
Area \( = \pi r^2 = \frac{22}{7} \times \frac{21}{4} \times \frac{21}{4} = \frac{22 \times 3 \times 21}{16} = 86.625 \text{ cm}^2 \)

Question. If the area of circle is \( 616 \text{ cm}^2 \), then its circumference is ...................... .
Answer: 88 cm
Explanation: Let \( r \) be the radius of circle
Area \( = 616 \text{ cm}^2 \)
\( \pi r^2 = 616 \)

\( \implies r^2 = \frac{616 \times 7}{22} = 28 \times 7 \)

\( \implies r^2 = 4 \times 7 \times 7 = (2 \times 7)^2 \)

\( \implies r = 14 \)
Circumference \( = 2\pi r = 2 \times \frac{22}{7} \times 14 = 88 \text{ cm} \)

Question. If the area of a semi-circular region is 308 sq cm, then its perimeter is .........................
Answer: 72 cm
Explanation: Area of semi-circular region \( = \frac{\pi}{2}(r^2) = 308 \)

\( \implies r = 14 \text{ cm} \)
So, perimeter \( = \pi r + 2r = 44 + 28 = 72 \text{ cm} \)

Question. Number of rounds that a wheel of diameter \( \frac{7}{11} \) metre will make in moving a distance of 2 km is ...............
Answer: 1000 rounds
Explanation: Number of rounds \( = \frac{2 \text{ km}}{2\pi \left(\frac{7}{22}\right) \text{ m}} = \frac{2000}{2} = 1000 \)