The questions on finding perimeters and areas related to circular figures are of great practical importance. This exercise contains problems on finding the perimeter and area of a circle, as well as the area enclosed by two concentric circles. To access detailed solutions to problems, the RD Sharma Solutions Class 10 is available for students for the purpose of clearing doubts. Downloadable format of the RD Sharma Solutions for Class 10 Maths Chapter 15 Areas Related to Circles Exercise 15.1 PDF are provided below.
RD Sharma Solutions for Class 10 Maths Chapter 15 Areas Related to Circles Exercise 15.1
Access RD Sharma Solutions for Class 10 Maths Chapter 15 Areas Related to Circles Exercise 15.1
1. Find the circumference and area of a circle of radius of 4.2 cm.
Solution:
Given,
Radius (r) = 4.2 cm
We know that
Circumference of a circle = 2πr
= 2 × (22/7) × 4.2 = 26.4 cm2
Area of a circle = πr2
= (22/7) x 4.22
= 22 x 0.6 x 4.2 = 55.44 cm2
Hence, the circumference and area of the circle are 26.4 cm2 and 55.44 cm2, respectively.
2. Find the circumference of a circle whose area is 301.84 cm2.
Solution:
Given,
Area of the circle = 301.84 cm2
We know that,
Area of a Circle = πr2 = 301.84 cm2
(22/7) × r2 = 301.84
r2 = 13.72 x 7 = 96.04
r = √96.04 = 9.8
So, the radius is = 9.8 cm.
Now, the circumference of a circle = 2πr
= 2 × (22/7) × 9.8 = 61.6 cm
Hence, the circumference of the circle is 61.6 cm.
3. Find the area of a circle whose circumference is 44 cm.
Solution:
Given,
Circumference = 44 cm
We know that,
Circumference of a circle = 2πr = 44 cm
2 × (22/7) × r = 44
r = 7 cm
Now, the area of a circle = πr2
= (22/7) × 7 × 7
= 154 cm2
Hence, the area of the circle = 154 cm2
4. The circumference of a circle exceeds the diameter by 16.8 cm. Find the circumference of the circle.
Solution:
Let the radius of the circle be r cm.
So, the diameter (d) = 2r [As radius is half the diameter]
We know that,
The ircumference of a circle (C) = 2πr
From the question,
The circumference of the circle exceeds its diameter by 16.8 cm.
C = d + 16.8
2πr = 2r + 16.8 [d = 2r]
2πr – 2r = 16.8
2r (π – 1) = 16.8
2r (3.14 – 1) = 16.8
r = 3.92 cm
Thus, radius = 3.92 cm
Now, the circumference of the circle (C) = 2πr
C = 2 × 3.14 × 3.92
= 24.64 cm
Hence, the circumference of the circle is 24.64 cm.
5. A horse is tied to a pole with 28 m long string. Find the area where the horse can graze.
Solution:
Given,
The length of the string (l) = 28 m
The area the horse can graze is the area of the circle with a radius equal to the length of the string.
We know that
The area of a circle = πr2
= (22/7) × 28 × 28 = 2464 m2
Hence, the area of the circle, which is the same as the area the horse can graze, is 2464 m2
6. A steel wire when bent in the form of a square, encloses an area of 121 cm2. If the same wire is bent in the form of a circle, find the area of the circle.
Solution:
Given,
Area of the square = a2 = 121 cm2
We know that
Area of the circle = πr2
Area of a square = a2
121 cm2 = a2
So, a = 11 cm
Thus, each side of the square = 11 cm
Now, the perimeter of the square = 4a
= 4 × 11 = 44 cm
From the question, it’s understood that
The perimeter of the square = The circumference of the circle
We know that the circumference of a circle (C) = 2πr
4a = 2πr
44 = 2(22/7)r
r = 7 cm
Now, the area of the circle = πr2
= (22/7) × 7 × 7 = 154 cm2
Hence, the area of the circle is 154 cm2.
7. The circumference of two circles is in the ratio of 2:3. Find the ratio of their areas.
Solution:
Let’s consider the radius of two circles, C1 and C2, to be r1 and r2.
We know that, the circumference of a circle (C) = 2πr
And their circumference will be 2πr1 and 2πr2.
So, their ratio is = r1: r2
Given, the circumference of two circles is in a ratio of 2: 3.
r1: r2 = 2: 3
Then, the ratios of their areas are given as
= 4/9
Hence, the ratio of their areas = 4: 9
8. The sum of the radii of two circles is 140 cm, and the difference in their circumference is 88 cm. Find the diameters of the circles.
Solution:
Let the radii of the two circles be r1 and r2.
And the circumferences of the two circles be C1 and C2.
We know that the circumference of the circle (C) = 2πr
Given,
Sum of radii of two circle, i.e., r1 + r2 = 140 cm … (i)
Difference in their circumference,
C1 – C2 = 88 cm
2πr1 – 2πr2 = 88 cm
2(22/7)(r1 – r2) = 88 cm
(r1 – r2) = 14 cm
r1 = r2 + 14….. (ii)
Substituting the value of r1 in equation (i), we have
r2 + r2 + 14 = 140
2r2 = 140 – 14
2r2 = 126
r2 = 63 cm
Substituting the value of r2 in equation (ii), we have
r1 = 63 + 14 = 77 cm
Therefore,
Diameter of circle 1 = 2r1 = 2 x 77 = 154 cm
Diameter of circle 2 = 2r2 = 2 × 63 = 126 cm
9. Find the radius of a circle whose circumference is equal to the sum of the circumferences of two circles of radii 15cm and 18cm.
Solution:
Given,
Radius of circle 1 = r1 = 15 cm
Radius of circle 2 = r2 = 18 cm
We know that the circumference of a circle (C) = 2πr
So, C1 = 2πr1 and C2 = 2πr2
Let the radius be r of the circle which is to be found and its circumference (C).
Now, from the question,
C = C1 + C2
2πr = 2πr1 + 2πr2
r = r1 + r2 [After dividing by 2π both sides]
r = 15 + 18
r = 33 cm
Thus, the radius of the circle = 33 cm
10. The radii of the two circles are 8 cm and 6 cm, respectively. Find the radius of the circle having its area equal to the sum of the areas of two circles.
Solution:
Given,
The Radii of the two circles are 6 cm and 8 cm
The area of circle with radius 8 cm = π (8)2 = 64π cm2
The area of circle with radius 6cm = π (6)2 = 36π cm2
The sum of areas = 64π + 36π = 100π cm2
Let the radius of the circle be x cm.
Area of the circle = 100π cm2 (from above)
πx2 = 100π
x = √100 = 10 cm
Therefore, the radius of the circle is 10 cm.
11. The radii of the two circles are 19 cm and 9 cm, respectively. Find the radius and area of the circle, which has circumferences equal to the sum of the circumference of two circles.
Solution:
Given,
Radius of circle 1 = r1 = 19 cm
Radius of circle 2 = r2 = 9 cm
We know that the circumference of a circle (C) = 2πr
So, C1 = 2πr1 and C2 = 2πr2
Let the radius be r of the circle which is to be found and its circumference (C).
Now, from the question,
C = C1 + C2
2πr = 2πr1 + 2πr2
r = r1 + r2 [After dividing by 2π both sides]
r = 19 + 9
r = 28 cm
Thus, the radius of the circle = 28 cm
So, the area of required circle = πr2 = (22/7) × 28 × 28 = 2464 cm2
12. The area of a circular playground is 22176 m2. Find the cost of fencing this ground at the rate of ₹50 per metre.
Solution:
Given,
Area of the circular playground = 22176 m2
And the cost of fencing per metre = ₹50
If the radius of the ground is taken as r.
Then, its area = πr2
πr2 = 22176
r2 = 22176(7/22) = 7056
Taking square root on both sides, we have
r = 84 m
We know that the fencing is done only on the circumference of the ground.
Circumference of the ground = 2πr = 2(22/7)84 = 528 m
So, the cost of fencing 528 m = ₹50 x 528 = ₹26400
Therefore, the cost of fencing the ground is ₹26,400.
13. The side of a square is 10 cm. Find the area of the circumscribed and inscribed circles.
Solution:
For the circumscribed circle,
Radius = diagonal of square/ 2
Diagonal of the square = side x √2
= 10√2 cm
Radius = (10 × 1.414)/ 2 = 7.07 cm
Thus, the radius of the circumcircle = 7.07 cm
Then, its area is = πr2 = (22/7) × 7.07 × 7.07 = 157.41 cm2
Therefore, the Area of the Circumscribed circle is 157.41 cm2
For the inscribed circle,
Radius = side of square/ 2
= 10/ 2 = 5 m
Then, its area is = πr2 = 3.14 × 5 × 5 = 78.5 cm2
Thus, the area of the circumscribed circle is 157.41 cm2, and the area of the inscribed circle is 78.5 cm2.
14. If a square is inscribed in a circle, find the ratio of the areas of the circle and the square.
Solution:
Let the side of the square be x cm which is inscribed in a circle.
Given,
Radius of circle (r) = 1/2 (diagonal of square)
= 1/2(x√2)
r = x/√2
We know that the area of the square = x2
And, the area of the circle = πr2
Therefore, the ratio of areas of the circle and the square = π : 2
15. The area of a circle inscribed in an equilateral triangle is 154 cm2. Find the perimeter of the triangle.
Solution:
Let the circle inscribed in the equilateral triangle be with a centre O and radius r.
We know that, Area of a Circle = πr2
But, given that area is 154 cm2.
(22/7) × r2 = 154
r2 = (154 x 7)/22 = 7 × 7 = 49
r = 7 cm
From the figure seen above, we infer that
At point M, the BC side is tangent and also, at point M, BM is perpendicular to OM.
We know that,
In an equilateral triangle, the perpendicular from the vertex divides the side into two halves.
BM = ½ x BC
Consider the side of the equilateral triangle is x cm.
After solving the above equation, we get
Therefore, the perimeter of the triangle is found to be 42√3 cm = 42(1.73) = 72.7 cm
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