MCQ Questions for Class 10 Maths Chapter 12 Areas Related to Circles

Areas Related to Circles Class 10 MCQ Questions with Answers

Question 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) \(R_{1}^{2}+R_{2}^{2}=R^{2}\)
(C) R1 + R2 < R
(D) \(R_{1}^{2}+R_{2}^{2}=R^{2}\)
Answer:
(B) \(R_{1}^{2}+R_{2}^{2}=R^{2}\)

Explanation:
According to the given condition, Area of circle =Area of first circle + Area of second circle
πR² = π \(R_{1}{ }^{2}\) + π\(R_{2}{ }^{2}\)
R² =\(R_{1}{ }^{2}\) + \(R_{2}{ }^{2}\).

Question 2.

If the sum of the circumferences of two circles with radii R₁ and R₂ is equal to the circumference of a circle of radius R, then:

(A) R₁ + R₂ = R
(B) R₁ + R₂ > R
(C) R₁ + R₂ < R
(D) Nothing definite can be said about the relation among R₁, R₂ and R
Answer:
(A) R₁ + R₂ = R

Explanation:
According to question, Circumference of circle
= Circumference of first circle + Circumference of second circle
2πR = 2πR₁ + 2πR₂
R = R₁ + R₂

Question 3.

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) Area of the circle > Area of the square

Explanation:
According to question, Circumference of a circle = Perimeter of square Let ‘r’ and ‘a’ be the radius of circle and side of square.
2πr = 4 a
\(\frac{22}{7}\)r = 2a
11r = 7a
r = \(\frac{7 a}{11}\) ….(i)
A₁ = πR²
From equation (i), we have
A₁ = π\(\left(\frac{7 a}{11}\right)^{2}\)
= \(\frac{22}{7}\left(\frac{49 a^{2}}{121}\right)\)
= \(\frac{14 a^{2}}{11}\)
A₂ = a²
From equation (ii) and (iii), we haveA1 = \(\frac{14}{11}\)A2
A₁ > A₂
Area equation is greater than the area of square.

Question 4.

Area of the largest triangle that can be inscribed in a semi-circle of radius ‘r’ units is:

(A) r² sq. units
(B) \(\frac{1}{2}\) r² sq. units
(C) 2r² sq units
(D) √2r² sq. units
Answer:
(A) r² sq. units

Explanation:
Take a point C on the circumference of the semi-circle and join it by the end points of diameter AB.

∠C = 90 [Angle in a semi – circle is right angle]
So ABC = \(\frac{1}{2}\) × AB × CD
= \(\frac{1}{2}\) × 2r × r = r² sQuestion units

Question 5.

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) 14 : 11

Explanation:
Let the radius of circle be ‘r’ and side of square be ‘a’.
According to given question,
Perimeter of circle = Perimeter of square
2πr = 4a
∴ a = \(\frac{\pi r}{2}\)
So, \(\frac{ Area of circle}{Area of square}\) = \(\frac{\pi r^{2}}{\left(\frac{\pi r}{2}\right)^{2}}\) [From equation (i)]
Solving equation (i), we get result as \(\frac{14}{11}\).

Question 6.

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

Explanation:
Area of first circular park whose diameter is 16 m,
= π\(\left(\frac{16}{2}\right)^{2}\)
= π\((8)^{2}\)
= 64π m²
Area of second circle park whose dimeter is 12 m,
= π\(\left(\frac{12}{2}\right)^{2}\)
= π(6)²
= 36π m²
According to question,
Area of single circlar park = Area of first circular park + Area of second circular park
πr² = 64π + 36π
πr² = 100π
r =10 m

Question 7.

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

(A) 36π cm²
(B) 18π cm²
(C) 12π cm²
(D) 9π cm²
Answer:
(D) 9π cm²

Explanation:

Given, side of square = 6 cm
Diameter of a circle, (d) = Side of square = 6 cm
Radius of a circle (r) = \(\frac{d}{2}\) = 6 = 3 cm.
Area of circle = πr²
π(3)² = 9π cm²

Question 8.

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

(A) 256 cm²
(B) 128 cm²
(C) 64 cm²
(D) 64 cm²
Answer:
(B) 128 cm²

Explanation:
Given, radius of circle, r =OC = 8 cm
Diameter of the circle
= AC = 2 × OC
= 2 × 8 = 16 cm
Which is equal to the diagonal of a square.

Let side of square be ‘a’a.
Using Pythagoras theorem,
AB² + BC² = AC2
a² + a² = 16²
2a² = 256
a² = 128 cm ²

Assertion and Reason Based MCQs

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is True

Question 1.

Assertion (A): 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 50 cm.
Reason (R): If the perimeter and the area of a circle are numerically equal, then the radius of the circle is 2 units.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
In the case of assertion:
Let r₁ = 24 cm and r₂ = 7 cm Area of first circle = π\(r_{1}^{2}\) = π(24)² = 576π cm²
Area of second circle = π\(r_{2}^{2}\) = π(7)² = 49π cm² cm²
According to question,
Area of circle = Area of first circle + Area of second circle
πR² = 576π + 49π
[where, R be radius of circle]
R² =625 = 25 cm
Diameter of a circle = 2R = 2 × 25 = 50 cm.
∴ Assertion is correct.
In case of reason:
Let the radius of the circle be r.
Circumference of circle = 2πr
Area of circle = πr²
Given that, the circumference of the circle and the area of the circle are equal. This implies,
2πr = πr²
r = 2
Therefore, the radius of the circle is 2 units.
∴ Reason is correct.
Hence, both assertion and reason are correct but reason is not correct explanation for assertion.

Question 2.

Assertion (A): In covering a distance s meter, a circular wheel of radius r meter makes \(\frac{s}{2 \pi r}\) revolution.
Reason (R): The distance travelled by a circular wheel of diameter d cm in one revolution is 2πd cm.

Answer:
(C) A is true but R is false

Explanation:
In case of assertion:
The distance covered in one revolution is 2πr, that is, its circumference.
∴ Assertion is correct In case of reason:
Because the distance travelled by the wheel in one revolution is equal to its circumference Circumference of wheel = π × diameter ,
= π × d
= πd
Hence the given answer in the question is incorrect.
∴Reason is incorrect.
Hence, assertion is correct and reason is incorrect.

Question 3.

Assertion (A): If circumferences of two circles are equal, then their areas will be equal.
Reason (R): If the areas of two circles are equal, then their circumferences are equal.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
In case of assertion:
If circumferences of two circles are equal, then their corresponding radii are equal. So, their areas will be equal.
∴ Assertion is correct.
In case of reason:
If areas of two circles are equal, then their corresponding radii are equal. So, their circumference will be equal.
∴ Reason is correct.
Hence, both assertion and reason are correct but reason is not the correct explanation for assertion.

Case – Based MCQs

Attempt any four sub-parts from each question. Each sub-part carries 1 mark.
I. Read the following text and answer the following question on the basis of the same:
A horse is tied to a peg at one comer of a square shaped grass field of sides 15 m by means of a 5 m long rope (see the given figure)

Question 1.

What is the area of the grass field?

(A) 225 m²
(B) 225 m
(C) 255 m²
(D) 15 m
Answer:
(A) 225 m²

Explanation:
Area of square = (side)²
= 15 × 15
= 225 m²

Question 2.

The area of that part of the field in which the horse can graze.

(A) 19.625 m²
(B) 19.265 m²
(C) 19 m²
(D) 78.5 m²
Answer:
(A) 19.625 m²

Explanation:
From the figure, it can be observe that the horse can graze a sector of 90° in a circle of 5 m radius.
Area that can be grazed by horse = Area of sector
= \(\frac{90^{\circ}}{360^{\circ}}\) × πr²
= \(\frac{1}{4}\) × 3 . 14 × 5 × 5
= 19 . 625 m²

Question 3.

The grazing area if the rope were 10 m long instead of 5 m.

(A) 7.85 m²
(B) 785 m²
(C) 225 m²
(D) 78.5 m²
Answer:
(D) 78.5 m²

Explanation:
Area that can be grazed by the horse when length of rope is 10 m long
= \(\frac{90^{\circ}}{360^{\circ}}\) × πr²
= \(\frac{1}{4}\) × 3 . 14 × 10 × 10
= 78.5 m²

Question 4.

The increase in the grazing area if the rope were 10 m long instead of 5 m.

(A) 58.758 m²
(B) 58.875 m²
(C) 58 m²
(D) 78.5 m²
Answer:
(B) 58.875 m²

Explanation:
Increase in grazing area
= (78.5 – 19.625) m²
= 58.875 m²

Question 5.

The given problem is based on which concept?

(A) Coordinate geometry
(B) Area related to circles
(C) Circle
(D) None of these
Answer:
(B) Area related to circles

II. Read the following text and answer the following question on the basis of the same:
In a workshop, brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the given figure.

Question 1.

What is the radius of the circle?

(A) \(\frac{35}{2} \mathrm{~mm}\)
(B) \(\frac{5}{2} \mathrm{~mm}\)
(C) 35 mm
(D) 10 mm
Answer:
(A) \(\frac{35}{2} \mathrm{~mm}\)

Explanation:
Radius of circle = \(\frac{\text { Diameter }}{2}\)
= \(\frac{35}{2} \mathrm{~mm}\)

Question 2.

What is the circumference of the brooch?

(A) 100 mm
(B) 110 mm
(C) 50 mm
(D) 10 mm
Answer:
(B) 110 mm

Explanation:
Circumference of brooch = 2πr
= 2 × \(\frac{22}{7}\) × \(\frac{35}{2}\)
= 110 mm

Question 3.

What is the total length of silver wire required?

(A) 528 mm
(B) 825 mm
(C) 285 mm
(D) 852 mm
Answer:
(C) 285 mm

Explanation:
Length of wire required
= 110 + 5 x 35
= 110 + 175
= 285 mm.

Question 4.

What is the area of the each sector of the brooch?

(A) \(\frac{385}{2} \mathrm{~mm}^{2}\)
(B) \(\frac{358}{4} \mathrm{~mm}^{2}\)
(C) \(\frac{585}{4} \mathrm{~mm}^{2}\)
(D) \(\frac{385}{4} \mathrm{~mm}^{2}\)
Answer:
(D) \(\frac{385}{4} \mathrm{~mm}^{2}\)

Explanation:
It can be observed from the figure that an angle of each 10 sectors of the circle is subtending at the centre of the circle.
∴Area of each sector = \(\frac{36}{360^{\circ}}\) × πr²
= \(\frac{1}{10}\) × \(\frac{22}{7}\) × \(\frac{35}{2}\) × \(\frac{35}{2}\)
= \(\frac{385}{4}\) mm2

Question 5.

The given problem is based on which mathematical concept ?

(A) Areas Related to circles
(B) Circles
(C) Construction
(D) none of these
Answer:
(A) Areas Related to circles

III. Read the following text and answer the following question on the basis of the same:
AREAS RELATED TO CIRCLES Pookalam is the flower bed or flower pattern designed during Onam in Kerala. It is similar as Rangoli in North India and Kolam in Tamil Nadu. During the festival of Onam, your school is planning to conduct a Pookalam competition. Your friend who is a partner in competition , suggests two designs given below.
Observe these carefully.

Design I: This design is made with a circle of radius 32 cm leaving equilateral triangle ABC in the middle as shown in the given figure.
Design II: This Pookalam is made with 9 circular design each of radius 7 cm.
Refer Design I:

Question 1.

The side of equilateral triangle is

(A) 12√3 cm
(B) 32√3 cm
(C) 48 cm
(D) 64 cm
Answer:
(B) 32√3 cm

Question 2.

The altitude of the equilateral triangle is

(A) 8 cm
(B) 12 cm
(C) 48 cm
(D) 52 cm
Answer:
(C) 48 cm

Refer Design II:

Question 3.

The area of square is

(A) 1264 cm²
(B) 1764 cm²
(C) 1830 cm²
(D) 1944 cm²
Answer:
(B) 1764 cm²

Explanation:
radius = 7 cm
diameter = 2 x 7 cm = 14 cm
side of square = 14 cm + 14 cm + 14 cm
= 42 cm.
Area of square = side²
= (42 cm)²
1764 cm²

Question 4.

Area of each circular design is

(A) 124 cm²
(B) 132 cm²
(C) 144 cm²
(D) 154 cm²
Answer:
(D) 154 cm²

Explanation:
radius = 7 cm
Area of each circular design = πr²
= \(\frac{22}{7}\) × 7 × 7
= 154 cm²

Question 5.

Area of the remaining portion of the square ABCD

(A) 378 cm²
(B) 260 cm²
(C) 340 cm²
(D) 278 cm²
Answer:
(A) 378 cm²

Explanation:
Area of 9 circular design = 9 × πr²
= 9 × \(\frac{22}{7}\) × 7 × 7
= 1386 cm²
Area of square = 1764 cm²
Area of remaining portion of square – Area of 9 circular design
= 1764 cm ² – 1386 cm²
= 378 cm²

IV. Read the following text and answer the following question on the basis of the same:
A Brooch A brooch is a small piece of jewellery which has a pin at the back so it can be fastened on a dress, blouse or coat. Designs of some brooch are shown below. Observe them carefully.



Design A: Brooch A is made with silver wire in the form of a circle with diameter 28 mm. The wire used for making 4 diameters which divide the circle into 8 equal parts.
Design B: Brooch b is made two colours i.e. Gold and silver. Outer part is made with Gold. The circumference of silver part is 44 mm and the gold part is 3 mm wide everywhere.
Refer to Design A

Question 1.

The total length of silver wire required is

(A) 180 mm
(B) 200 mm
(C) 250 mm
(D) 280 mm
Answer:
(B) 200 mm

Explanation:
Diameter = 28 mm
radius = 14 mm
Total length of wire = length of 4 diameter + circumference of circle.
= 4 × 28 + 2πr²
= 112 + 2 × \(\frac{22}{7}\) × 14
= 112 + 88
= 200 mm

Question 2.

The area of each sector of the brooch is

(A) 44 mm2
(B) 52 mm2
(C) 77 mm2
(D) 68 mm2
Answer:
(C) 77 mm2

Explanation:
Area of each sector of Brooch
= \(\frac{1}{8}\) × Area of Brooch
= \(\frac{1}{8}\) × πr²
= \(\frac{1}{8}\) x \(\frac{22}{7}\) × 14 × 14
= 77 mm²

Refer to Design B

Question 3.

The circumference of outer part (golden) is

(A) 48.49 mm
(B) 82.2 mm
(C) 72.50 mm
(D) 62.86 mm
Answer:
(D) 62.86 mm

Question 4.

The difference of areas of golden and silver parts is

(A) 18π
(B) 44π
(C) 51π
(D) 64π
Answer:
(C) 51π

Question 5.

A boy is playing with brooch B. He makes revolution with it along its edge. How many complete revolutions must it take to cover 80 mm ?

(A) 2
(B) 3
(C) 4
(D) 5
Answer:
(C) 4

Explanation:
Circumference of silver part of Brooch
= 44 cm
2πr = 44 mm
2 × \(\frac{22}{7}\) × r = 44
r = 7 mm.
radius of whole Brooch
= 7 mm + 8 mm
= 10 mm.
Circumference of outer edge
= 2πr
= 2 × \(\frac{22}{7}\) × 10
= \(\frac{440}{7}\) mm
let the number of revolutions = n
Now, According to question,
n . 2πr = 80π
n . \(\frac{440}{7}\) = 80π
n . \(\frac{440}{7}\) = 80 x \(\frac{22}{7}\)
n = 4

MCQ Questions for Class 10 Maths with Answers