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 cm^{2}

Area of a circle = πr^{2}

= (22/7) x 4.2^{2}

= 22 x 0.6 x 4.2 = 55.44 cm^{2}

Hence, the circumference and area of the circle are 26.4 cm^{2} and 55.44 cm^{2}, respectively.

**2. Find the circumference of a circle whose area is 301.84 cm ^{2}. **

**Solution:**

Given,

Area of the circle = 301.84 cm^{2 }

We know that,

Area of a Circle = πr^{2 }= 301.84 cm^{2 }

(22/7) × r^{2} = 301.84

r^{2} = 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 = πr^{2}

= (22/7) × 7 × 7

= 154 cm^{2}

Hence, the area of the circle = 154 cm^{2}

**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 = πr^{2}

= (22/7) × 28 × 28 = 2464 m^{2}

Hence, the area of the circle, which is the same as the area the horse can graze, is 2464 m^{2}

**6. A steel wire when bent in the form of a square, encloses an area of 121 cm ^{2}. If the same wire is bent in the form of a circle, find the area of the circle. **

**Solution:**

Given,

Area of the square = a^{2} = 121 cm^{2}

We know that

Area of the circle = πr^{2}

Area of a square = a^{2}

121 cm^{2} = a^{2}

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 = πr^{2}

= (22/7) × 7 × 7 = 154 cm^{2}

Hence, the area of the circle is 154 cm^{2}.

**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, C_{1} and C_{2}, to be r_{1} and r_{2}.

We know that, the circumference of a circle (C) = 2πr

And their circumference will be 2πr_{1} and 2πr_{2}.

So, their ratio is = r_{1}: r_{2}

Given, the circumference of two circles is in a ratio of 2: 3.

r_{1}: r_{2} = 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 r_{1} and r_{2}.

And the circumferences of the two circles be C_{1} and C_{2}.

We know that the circumference of the circle (C) = 2πr

Given,

Sum of radii of two circle, i.e., r_{1} + r_{2} = 140 cm … (i)

Difference in their circumference,

C_{1} – C_{2} = 88 cm

2πr_{1} – 2πr_{2 }= 88 cm

2(22/7)(r_{1} – r_{2}) = 88 cm

(r_{1} – r_{2)} = 14 cm

r_{1 =} r_{2} + 14…..** **(ii)** **

Substituting the value of r_{1 }in equation (i)**, **we have

r_{2 }+ r_{2} + 14 = 140

2r_{2} = 140 – 14

2r_{2 }= 126

r_{2} = 63 cm

Substituting the value of r_{2 }in equation (ii)**, **we have

r_{1 }= 63 + 14 = 77 cm

Therefore,

Diameter of circle 1 = 2r_{1 }= 2 x 77 = 154 cm

Diameter of circle 2 = 2r_{2 }= 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 = r_{1} = 15 cm

Radius of circle 2 = r_{2} = 18 cm

We know that the circumference of a circle (C) = 2πr

So, C_{1} = 2πr_{1 }and C_{2} = 2πr_{2}

Let the radius be r of the circle which is to be found and its circumference (C).

Now, from the question,

C = C_{1} + C_{2}

2πr = 2πr_{1} + 2πr_{2}

r = r_{1} + r_{2 }[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 cm^{2}

The area of circle with radius 6cm = π (6)^{2} = 36π cm^{2}

The sum of areas = 64π + 36π = 100π cm^{2}

Let the radius of the circle be x cm.

Area of the circle = 100π cm^{2} (from above)

πx^{2 }= 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 = r_{1} = 19 cm

Radius of circle 2 = r_{2} = 9 cm

We know that the circumference of a circle (C) = 2πr

So, C_{1} = 2πr_{1 }and C_{2} = 2πr_{2}

Let the radius be r of the circle which is to be found and its circumference (C).

Now, from the question,

C = C_{1} + C_{2}

2πr = 2πr_{1} + 2πr_{2}

r = r_{1} + r_{2 }[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 = πr^{2} = (22/7) × 28 × 28 = 2464 cm^{2}

**12. The area of a circular playground is 22176 m ^{2}. Find the cost of fencing this ground at the rate of ₹50 per metre. **

**Solution: **

Given,

Area of the circular playground = 22176 m^{2}

And the cost of fencing per metre = ₹50

If the radius of the ground is taken as r.

Then, its area = πr^{2}

πr^{2} = 22176

r^{2} = 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 = πr^{2} = (22/7) × 7.07 × 7.07 = 157.41 cm^{2 }

Therefore, the Area of the Circumscribed circle is 157.41 cm^{2 }

For the inscribed circle,

Radius = side of square/ 2

= 10/ 2 = 5 m

Then, its area is = πr^{2} = 3.14 × 5 × 5 = 78.5 cm^{2 }

Thus, the area of the circumscribed circle is 157.41 cm^{2}, and the area of the inscribed circle is 78.5 cm^{2}.

**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 = x^{2}

And, the area of the circle = πr^{2}

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 cm ^{2}. 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 = πr^{2}

But, given that area is 154 cm^{2}.

(22/7) × r^{2} = 154

r^{2} = (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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