1. If the sum of the areas of two circles with radii R₁ and R₂ is equal to the area of a circle of radius R, then

(A) R₁ + R₂ = R

(B) R₁² + R₂² = R²

(C) R₁ + R₂ < R

(D) R₁² + R₂² < R²

Answer: (B)

Explanation: According to given condition,

Area of circle = Area of first circle + Area of second circle

2. If the circumference of a circle and the perimeter of a square are equal, then

(A) Area of the circle = Area of the square

(B) Area of the circle > Area of the square

(C) Area of the circle < Area of the square

(D) Nothing definite can be said about the relation between the areas of the circle and square.

Answer:  (B)

Explanation: According to given condition

Circumference of a circle = Perimeter of square

Hence Area of the circle > Area of the square

3. Area of the largest triangle that can be inscribed in a semi-circle of radius r units, in square units is:

(A) r²

(B) 1/2r²

(C) 2 r²

(D) √2r²

Answer:  (A)

Explanation: The triangle inscribed in a semi-circle will be the largest when the perpendicular height of the triangle is the same size as the radius of the semi-circle.

Consider the following figure:

4. 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) 7:22

(D) 11:14

Answer: (B)

Explanation: Perimeter of circle = Perimeter of square

5. 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)

Explanation: Area of first circular park, whose diameter is 16m

= πr² = π (16/2)² = 64π m²

Area of second circular park, whose diameter is 12m

= πr² = π (12/2)² = 36π m²

According to question,

Area of new circular park =

6. The area of the circle that can be inscribed in a square of side 6 cm is

(A) 36 π cm²

(B) 18 π cm²

(C) 12 π cm²

(D) 9 π cm²

Answer:  (D)

Explanation: Given,

Side of square = 6 cm

Diameter of a circle = side of square = 6cm

Therefore, Radius of circle = 3cm

Area of circle

= πr²

= π (3)²

= 9π cm²

7. The area of the square that can be inscribed in a circle of radius 8 cm is

(A) 256 cm²

(B) 128 cm²

(C) 642 cm²

(D) 64 cm²

Answer:  (B)

Explanation: Radius of circle = 8 cm

Diameter of circle = 16 cm = diagonal of the square

Therefore side of square = diagonal/√2

= 16/√2

Therefore, are of square is = (side)² = (16/√2)²

= 256/2

= 128 cm²

8. 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)

Explanation: According to question,

Circumference of circle = Circumference of first circle + Circumference of second circle

πD = πd₁ + πd₂

D = 36 + 20

D = 56cm

9. 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 respectively, is

(A) 31 cm

(B) 25 cm

(C) 62 cm

(D) 50 cm

Answer:  (D)

Explanation: According to question

Therefore diameter = 2 × 25 = 50cm

10. If the length of an arc of a circle of radius r is equal to that of an arc of a circle of radius 2r, then

(A) the angle of the corresponding sector of the first circle is double the angle of the corresponding sector of the other circle.

(B) the angle of the corresponding sector of the first circle is equal the angle of the corresponding sector of the other circle.

(C) the angle of the corresponding sector of the first circle is half the angle of the corresponding sector of the other circle.

(D) the angle of the corresponding sector of the first circle is 4 times the angle of the corresponding sector of the other circle.

Answer:  (A)

Explanation: According to Question,

11. The wheel of a motor cycle is of radius 35 cm. How many revolutions per minute must the wheel make so as to keep a speed of 66 km/h?

(A) 300

(B) 400

(C) 450

(D) 500

Answer: (D)

Explanation:

12. A cow is tied with a rope of length 14 m at the corner of a rectangular field of dimensions 20m × 16m, then the area of the field in which the cow can graze is:

(A) 154 m²

(B) 156 m²

(C) 158 m²

(D) 160 m²

Answer:  (A)

Explanation:Figure according to question is:

Area of the field in which cow can graze= Area of a sector AFEG

= (θ/360) X πr²

= (90/360) X π (14)²

= (1/4) X (22/7) X 196

= 154 m²

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13. The area of the shaded region in Fig., where arcs drawn with centres P, Q, Rand S intersect in pairs at mid-points A, B, C and D of the sides PQ, QR, RS and SP, respectively of a square PQRS, is:

(A) 25.25 cm²

(B) 27.45 cm²

(C) 29.65 cm²

(D) 30.96 cm²

Answer:  (D)

Explanation:

14. Area of a sector of central angle 120° of a circle is 3π cm². Then the length of the corresponding arc of this sector is:

(A) 5.8cm

(B) 6.1cm

(C) 6.3cm

(D) 6.8cm

Answer:  (C)

Explanation:

Given that

Area of a sector of central angle 120° of a circle is 3π cm²

15. A round table cover has six equal designs as shown in the figure. If the radius of thecover is 28 cm, then the cost of making the design at the rate of Rs. 0.35 per cm² is:

(A) Rs.146.50

(B) Rs.148.75

(C) Rs.152.25

(D) Rs.154.75

Answer:  (B)

Explanation: The area of the hexagon will be equal to six equilateral triangles with each side equal to the radius of circle.

Area of given hexagon = Area of 6 equilateral triangles.

= 6 X (√3/4) X (side)²

= 6 X (√3/4) X (28)²

= 1999.2 cm²                           (Taking √3 = 1.7)        

Area of circle = πr²

= π × 28²

= 2464 cm²

So, area of designed portion = 2464 – 1999.2 = 464.8 cm²

Cost of making design = 464.8 × 0.35

= Rs. 162.68

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Quick Revision Notes : Areas Related to Circles

NCERT Solution : Areas Related to Circles

MCQs: Areas Related to Circles

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