NCERT Exemplar Solutions for Class 10 Maths Chapter 11 Area Related to Circles Exercise 11.1

Chapter 11 Area Related to Circles NCERT Exemplar Solutions Exercise 11.1 Class 10 Maths

Chapter Name	NCERT Maths Exemplar Solutions for Chapter 11 Area Related to Circles Exercise 11.1
Book Name	NCERT Exemplar for Class 10 Maths
Other Exercises	Exercise 11.2 Exercise 11.3 Exercise 11.4
Related Study	NCERT Solutions for Class 10 Maths

Exercise 11.1 Solutions

Multiple Choice Questions

Choose the correct answer from the given four options:

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

Solution

(B) R₁² + R₂² = R² 
Justification : 
We have, 
Area of circle  = Area of first circle + Area of second circle 
πR² = πR₁² + πR₂² 

⇒ R² = R₁² + R₂² 
Option B is correct. 

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 & R.

Solution

(A) R₁ + R₂ = R
Justification : 

We have, 

Circumference of circle with radius R = Circumference of first circle with radius R₁ +

Circumference of second circle with radius R₂ 
πR = πR₁ + πR₂ 
⇒ R = R₁ + R₂ 
Option A is correct.

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 & square.

Solution

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

Justification:

We have,

Circumference of a circle = Perimeter of square

Let r be the radius of the circle and a be the side of square.

From the given condition,

2πr = 4a

⇒ 11r = 7a
⇒ a = (11/7)r 
⇒ r = (7/11)a 
Now, 
Area of circle = A₁ = πr² and 
Area of square = A₂ = a² 
From equation (i), we have 

We get, 
A₁ > A₂ 
Therefore, Area of the circle > Area of the square. 
Option B is correct. 

4. Area of the largest triangle that can be inscribed in a semi-circle of radius r units is

(A) r² sq. units 

(B) (1/2)r² sq. units

(C) 2r² sq. units

(D) √2r² sq. units

Solution

(A) r² sq. units 

Justification: 

The largest triangle that can be inscribed in a semi-circle of radius r units is the triangle having its base as the diameter of the semi-circle and the two other sides are taken by considering a point C on the circumference of the semi-circle and joining it by the end points of diameter A and B.

∠ C = 90° (by the properties of circle)

ΔABC is right angled triangle with base as diameter AB of the circle and height be CD.

Height of the triangle = r

Area of largest ΔABC = (1/2) × Base × Height 
= (1/2) × AB × CD
= (1/2) × 2r × r
= r²  sq. units
Option A is correct.

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

Solution

(B) 14 : 11

Justification:

Let r be the radius of the circle and a be the side of the square.

We have,

Perimeter of a circle = Perimeter of a square

2πr = 4a
⇒ a = πr/2 
Area of the circle = πr² and 
Area of the square = a² 
Now, 
Ratio of their areas = (Area of circle)/(Area of square)

Option B is correct. 

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

Solution

(A) 10 m 

We have,

Area of first Circular Park, whose diameter is 16 m,
= πr²/2 
= π(16/2)² 
= 64π sq. metre 

Area of second Circular Park, whose diameter is 12m, 
= πr²/2 
= π(12/2)² 
= 36π sq. metre 

According to the question, 
Area of single circular park = area of 1st circular park + area of 2nd circular park 

πR² = 36π + 64π
⇒ R² = 100
⇒ R = 10 m

7. The area of the circle that can be inscribed in a square of side 6 cm is 
(A) 36 π cm² 
(B) 18π cm² 
(C) 12π cm² 
(D) 9π cm² 

Solution

(D) 9π cm² 

We have,

Side of square = 6cm

Side of square = diameter of circle = 6cm

So,

Radius of the circle = 3cm

Area of circle = πr² 

= π3² 

= 9π cm² 

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√2 cm² 
(D) 64 cm² 

Solution

(B) 128 cm² 

We have,

Radius of circle, r = OC = 8cm.

Diameter of the circle = AC

= 2 × OC

= 2 × 8

= 16 cm

Which is equal to the diagonal of a square.

As square is inscribed in circle,

Diameter of the circle = Diagonal of a square

So,

Area of square,
= (Diagonal)²/2
= 16²/2
= 256/2
= 128 cm² 

9. The radius of a circle whose circumference is equal to the sum of the circumferences of the two circles of diameters 36cm and 20 cm is

(A) 56 cm

(B) 42 cm

(C) 28 cm

(D) 16 cm

Solution

(C) 28 cm 

Circumference of first circle  = 2πr 

= πd₁ 
= 36π cm  [given, d₁ = 36 cm]
Circumference of second circle = πd₂ 
= 20π  cm  [given, d₂ = 20 cm]
We have, 
Circumference of circle  = Circumference of first circle + Circumference of second circle 
πD = 36π + 20π
⇒ D = 56 cm 

Required radius of circle = 28 cm 

10. 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

Solution

(D) 50 cm 

Let, 
r₁ = 24 cm 

r₂ = 7 cm 

Area of first circle = πr₁² 
= π(24)² 
= 576 π cm² 
Area of second circle = πr₂² 
= π(7)² 
= 49 π cm² 
We have, 
Area of circle = Area of first circle + Area of second circle 
πR² = 576π + 49π
⇒ R = 25 cm 

Required diameter of circle = 2 × 25 = 50 cm