RBSE Solutions for Class 10 Maths Chapter 15 Circumference and Area of a Circle Ex 15.1 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 15 Circumference and Area of a Circle Exercise 15.1.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 15 |

Chapter Name |
Circumference and Area of a Circle |

Exercise |
Exercise 15.1 |

Number of Questions Solved |
10 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 10 Maths Chapter 15 Circumference and Area of a Circle Ex 15.1

Question 1.

The radius of a circle is 3.5 cm. Find (RBSESolutions.com) its area and circumference.

Solution :

Given :

Radius of circle (r) = 3.5 cm

Circumference of circle = 2πr

= 2 × \(\frac { 22 }{ 7 }\) × 3.5

= 22 cm

Area of circle = πr^{2}

= \(\frac { 22 }{ 7 }\) × (3.5) × (3.5)

= 38.5 cm^{2}

Thus, circumference of circle is 22 cm and area is 38.5 cm^{2}

Question 2.

The circumference of a circle Is 44 m. Find (RBSESolutions.com) area of circle.

Solution :

Given

Circumference = 44 m

Let radius of circle = r m

Thus, 2πr = 44

⇒ 2 × \(\frac { 22 }{ 7 }\) × r = 44

⇒ r = \(\frac { 44\times 7 }{ 2\times 22 } \)

= 7 m

Area of circle = πr^{2}

= \(\frac { 22 }{ 7 }\) × 7 × 7

= 154 m^{2}

Thus, area of circle is 154 m^{2}

Question 3.

The radius of a semicircular plot is 21 m. Find (RBSESolutions.com) its area and perimeter.

Solution :

Given :

Radius of semi-circular plot

(r) = 21 m

Area = \(\frac { 1 }{ 2 }\)πr^{2}

= \(\frac { 1 }{ 2 }\) × \(\frac { 22 }{ 7 }\) × 21 × 21

= 693 m^{2}

and perimeter = \(\frac { 1 }{ 2 }\) × 2πr + 2r

= πr + 2r

= \(\frac { 22 }{ 7 }\) × 21 + 2 × 21

= 66 + 42

= 108 m

Thus, area of plot ¡s 693 m^{2} and perimeter is 108 m

Question 4.

The wheel of scooter covered 88 m distance in 100 revolution. Find (RBSESolutions.com) the radius of wheel

Solution :

Given :

In 100 revolutions, wheel covered a distance of 88 m.

Distance covered by wheel in 1 rotation = circumference of wheel

= \(\frac { 14 }{ 100 }\) m

= \(\frac { 14 }{ 100 }\) × 100 cm

= 14 cm

Thus, radius of wheel = 14 cm

Question 5.

The area of a circular plate is 154 cm^{2}. Find (RBSESolutions.com) its circumference.

Solution :

Given :

Area of circular = 154 cm^{2}

Lei radius of plate i.s r cm, then

πr^{2} = 154

Circumference of circular plate = 2πr

= 2 × \(\frac { 22 }{ 7 }\) × 7

= 44 cm

Thus, Circumference of circular plate = 44 cm

Question 6.

The circumference of circle is equal to perimeter of a (RBSESolutions.com) square. If area of square is 484 sq m, then find the area of circle.

Solution:

Let side of square = x m

then perimeter of square 4 × x

and area of square = x^{2}

According to question

Area of square =484 sq m

x^{2} = 484

x = \(\sqrt { 484 }\)

= 22 m

Perimeter of square = 4x = 4 × 22 = 88 m

Let radius of circle is r

circumference of circle = perimeter of square

⇒ 2πr = 88

⇒ 2 × \(\frac { 22 }{ 7 }\) × r = 88

r = \(\frac { 88\times 7 }{ 2\times 22 } \) = 14 m

Thus, area of the circle = πr^{2}

= \(\frac { 22 }{ 7 }\) × 14 × 14 = 616 sq m.

Thus, area of the circle = 616 sq m.

Question 7.

The cost of fencing a circular field at (RBSESolutions.com) the rate 24 per meter is ₹ 5280 and rate of ploughing is ₹ 0.50 per meter. Find the cost of ploughing the field.

Solution :

Cost of fencing at the rate ₹ 24 per m = ₹ 5280

Let, radius of circular field = r m

Thus, circumference of circular field

Thus, radius of circular field = 35 m

Area of circular field = πr^{2}

= \(\frac { 22 }{ 7 }\) × 35 × 35 = 3850 sq. m.

∵ cost of ploughing the field 1 m^{2} = ₹ 0.50

∴ cost of ploughing the field 3850 sq. m.

= 0.50 × 3850 = ₹ 1925

Thus, cost of ploughing the plot = ₹ 1925

Question 8.

The radius of a circular grass (RBSESolutions.com) field in 35 m. There is 7 m wide ramp around it. Find the area of ramp.

Solution :

Given :

Radius of circular grass field = 35 m.

Width of ramp = 7 m.

R = 35 + 7 = 42 m

Thus area of ramp

Thus, area of ramp is 1694 sq m

Question 9.

The area between two (RBSESolutions.com) concentric circles will be :

(A) πR^{2}

(B) π (R + r)(R – r)

(C) π(R^{2} – r)

(D) None of these

Solution :

= πR^{2} – πr^{2}

= π(R^{2} – r^{2})

= π (R + r)(R – r)

Thus, option (B) is correct.

Question 10.

Radius of two concentric circles (RBSESolutions.com) are 4 cm and 3 cm respectively. Then area bounded by these circles will be:

(A) 22 cm^{2}

(B) 12 cm^{2}

(C) 32 cm^{2}

(D) 18 cm^{2}

Solution :

Given : Radius of two concentre circle are 4 cm and 3 cm.

Thus, R= 4 cm, r = 3 cm

∴ Area between two concentric circles

= π(R^{2} – r^{2})

= \(\frac { 22 }{ 7 }\) [(4)^{2} – (3)^{2}]

= \(\frac { 22 }{ 7 }\) × [(4 + 3) (4 – 3)]

= \(\frac { 22 }{ 7 }\) × 7 = 22 cm^{2}

Thus option (A) is correct.

We hope the given RBSE Solutions for Class 10 Maths Chapter 15 Circumference and Area of a Circle Ex 15.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 15 Circumference and Area of a Circle Exercise 15.1, drop a comment below and we will get back to you at the earliest.

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