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 
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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 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}
(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}
Solution
(B) R_{1}^{2} + R_{2}^{2} = R^{2}
Justification :
We have,
Area of circle = Area of first circle + Area of second circle
Ï€R^{2} = Ï€R_{1}^{2} + Ï€R_{2}^{2}
Justification :
We have,
Area of circle = Area of first circle + Area of second circle
Ï€R^{2} = Ï€R_{1}^{2} + Ï€R_{2}^{2}
⇒ R^{2} = R_{1}^{2} + R_{2}^{2}
Option B is correct.
Option B is correct.
2. 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
(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} & R.
(C) R_{1} + R_{2} < R
(D) Nothing definite can be said about the relation among R_{1} , R_{2} & R.
Solution
(A) R_{1} + R_{2} = R
Justification :
Justification :
We have,
Circumference of circle with radius R = Circumference of first circle with radius R_{1} +
Circumference of second circle with radius R_{2}
Ï€R = Ï€R_{1} + Ï€R_{2}
⇒ R = R_{1} + R_{2}
Option A is correct.
Ï€R = Ï€R_{1} + Ï€R_{2}
⇒ R = R_{1} + R_{2}
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,
2Ï€r = 4a
⇒ 11r = 7a
⇒ a = (11/7)r
⇒ r = (7/11)a
Now,
Area of circle = A_{1} = Ï€r^{2} and
Area of square = A_{2} = a^{2}
From equation (i), we have
We get,
A_{1} > A_{2}
Therefore, Area of the circle > Area of the square.
Option B is correct.
⇒ a = (11/7)r
⇒ r = (7/11)a
Now,
Area of circle = A_{1} = Ï€r^{2} and
Area of square = A_{2} = a^{2}
From equation (i), we have
We get,
A_{1} > A_{2}
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 semicircle of radius r units is
(A) r^{2} sq. units
(B) (1/2)r^{2} sq. units
(C) 2r^{2} sq. units
(D) √2r^{2} sq. units
Solution
(A) r^{2} sq. units
Justification:
The largest triangle that can be inscribed in a semicircle of radius r units is the triangle having its base as the diameter of the semicircle and the two other sides are taken by considering a point C on the circumference of the semicircle and joining it by the end points of diameter A and B.
∠ C = 90° (by the properties of circle)
Î”ABC is right angled triangle with base as diameter AB of the circle and height be CD.
Height of the triangle = r
Area of largest Î”ABC = (1/2) × Base × Height
= (1/2) × AB × CD
= (1/2) × 2r × r
= r^{2} sq. units
Option A is correct.
= (1/2) × AB × CD
= (1/2) × 2r × r
= r^{2} 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
⇒ R^{2} = 100
⇒ R = 10 m
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
2Ï€r = 4a
⇒ a = Ï€r/2
Area of the circle = Ï€r^{2} and
Area of the square = a^{2}
Now,
Ratio of their areas = (Area of circle)/(Area of square)
Option B is correct.
⇒ a = Ï€r/2
Area of the circle = Ï€r^{2} and
Area of the square = a^{2}
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
Ï€R^{2} = 36Ï€ + 64Ï€We have,
Area of first Circular Park, whose diameter is 16 m,
= Ï€r^{2}/2
= Ï€(16/2)^{2}
= 64Ï€ sq. metre
= Ï€r^{2}/2
= Ï€(16/2)^{2}
= 64Ï€ sq. metre
Area of second Circular Park, whose diameter is 12m,
= Ï€r^{2}/2
= Ï€(12/2)^{2}
= 36Ï€ sq. metre
= Ï€r^{2}/2
= Ï€(12/2)^{2}
= 36Ï€ sq. metre
According to the question,
Area of single circular park = area of 1st circular park + area of 2nd circular park
Area of single circular park = area of 1st circular park + area of 2nd circular park
⇒ R^{2} = 100
⇒ R = 10 m
7. The area of the circle that can be inscribed in a square of side 6 cm is
(A) 36 Ï€ cm^{2}
(B) 18Ï€ cm^{2}
(C) 12Ï€ cm^{2}
(D) 9Ï€ cm^{2}
(A) 36 Ï€ cm^{2}
(B) 18Ï€ cm^{2}
(C) 12Ï€ cm^{2}
(D) 9Ï€ cm^{2}
Solution
(D) 9Ï€ cm^{2}
We have,
Side of square = 6cm
Side of square = diameter of circle = 6cm
So,
Radius of the circle = 3cm
Area of circle = Ï€r^{2}
= Ï€3^{2}
= 9Ï€ cm^{2}
8. 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) 64√2 cm^{2}
(D) 64 cm^{2}
(B) 128 cm^{2}
(C) 64√2 cm^{2}
(D) 64 cm^{2}
Solution
(B) 128 cm^{2}
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
= 16^{2}/2
= 256/2
= 128 cm^{2}
= (Diagonal)^{2}/2
= 16^{2}/2
= 256/2
= 128 cm^{2}
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 = 2Ï€r
= Ï€d_{1}
= 36Ï€ cm [given, d_{1} = 36 cm]
Circumference of second circle = Ï€d_{2}
= 20Ï€ cm [given, d_{2} = 20 cm]
We have,
Circumference of circle = Circumference of first circle + Circumference of second circle
Ï€D = 36Ï€ + 20Ï€
⇒ D = 56 cm
= 36Ï€ cm [given, d_{1} = 36 cm]
Circumference of second circle = Ï€d_{2}
= 20Ï€ cm [given, d_{2} = 20 cm]
We have,
Circumference of circle = Circumference of first circle + Circumference of second circle
Ï€D = 36Ï€ + 20Ï€
⇒ 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
(D) 50 cm
Let,
r_{1} = 24 cm
r_{1} = 24 cm
r_{2} = 7 cm
Area of first circle = Ï€r_{1}^{2}
= Ï€(24)^{2}
= 576 Ï€ cm^{2}
Area of second circle = Ï€r_{2}^{2}
= Ï€(7)^{2}
= 49 Ï€ cm^{2}
We have,
Area of circle = Area of first circle + Area of second circle
Ï€R^{2} = 576Ï€ + 49Ï€
⇒ R = 25 cm
= Ï€(24)^{2}
= 576 Ï€ cm^{2}
Area of second circle = Ï€r_{2}^{2}
= Ï€(7)^{2}
= 49 Ï€ cm^{2}
We have,
Area of circle = Area of first circle + Area of second circle
Ï€R^{2} = 576Ï€ + 49Ï€
⇒ R = 25 cm
Required diameter of circle = 2 × 25 = 50 cm