Practicing the Class 10 Maths MCQs on NCERT Chapter Areas Related to Circles is one of the best ways to prepare for the CBSE Class 10 Board Exam 2020.All these questions are provided with answers and their detailed explanation. Questions are also available in PDF format. By practicing these maths questions, students can not only revise important topics but also improve their speed and accuracy which can help them perform well in the exam.

**Check below the solved MCQs from Class 10 Maths Chapter 12 Areas Related to Circles:**

**1.** 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)

**Explanation:** According to given condition,

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

πR^{2} = πR_{1}^{2} + πR_{2}^{2}

R^{2} = R_{1}^{2} + R_{2}^{2}

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**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) r^{2}

(B) 1/2r^{2}

(C) 2 r^{2}

(D) √2r^{2}

**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:

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**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

= πr^{2} = π (16/2)^{2} = 64π m^{2}

Area of second circular park, whose diameter is 12m

= πr^{2} = π (12/2)^{2} = 36π m^{2}

According to question,

Area of new circular park =

πR^{2} = (64π + 36π) m^{2}

πR^{2} = 100π m^{2}

R = 10m

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

(A) 36 π cm^{2}

(B) 18 π cm^{2}

(C) 12 π cm^{2}

(D) 9 π cm^{2}

**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

= πr^{2}

= π (3)^{2}

= 9π cm^{2}

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

(A) 256 cm^{2}

(B) 128 cm^{2}

(C) 642 cm^{2}

(D) 64 cm^{2}

**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 cm^{2}

**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 = πd_{1} + πd_{2}

D = 36 + 20

D = 56cm

So, Radius = 56/2 = 28cm

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**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 2*r*, 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 m^{2}

(B) 156 m^{2}

(C) 158 m^{2}

(D) 160 m^{2}

**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 πr^{2}

= (90/360) X π (14)^{2}

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

= 154 m^{2}

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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 cm^{2}

(B) 27.45 cm^{2}

(C) 29.65 cm^{2}

(D) 30.96 cm^{2}

**Answer: (D)**

**Explanation:**

**CBSE Class 10th Maths Chapter-wise Important Formulas, Theorems & Properties**

**14.** Area of a sector of central angle 120° of a circle is 3π cm^{2}. 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 cm^{2}

**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 cm^{2} 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 cm^{2} (Taking √3 = 1.7)

Area of circle = πr^{2}

= π × 28^{2}

= 2464 cm^{2}

So, area of designed portion = 2464 – 1999.2 = 464.8 cm^{2}

Cost of making design = 464.8 × 0.35

= Rs. 162.68

**Students may download all of the above questions in PDF format so that they can practice them any time in offline mode as well. Link to download is given below: **

**To check the important MCQs fromall other chapters of CBSE Class 10 Maths, go to the following link:**

**Also check important resources for the preparation of CBSE Class 10 Board Exam 2020:**

When only few days are left for the board exams to begin, class 10 students would be seeking for some good and reliable resources which can help them revise their syllabus effectively in a short time. We have prepared some articles to bring you all necessary information and important preparation resources for the upcoming Class 10 CBSE Exams. Links to all these articles are provided below:

**CBSE Class 10 Exam Pattern 2020: All Subjects****CBSE Class 10 Syllabus for Board Exam 2020: All Subjects****CBSE Class 10 Sample Papers with Marking Scheme 2020: All Subjects****CBSE Class 10 Question Papers 2019 with Solutions****CBSE Class 10 NCERT Exemplar Problems, Solutions for Maths & Science****CBSE Class 10 NCERT New Textbooks & NCERT Solutions: All Subjects**