RBSE Solutions for Class 10 Maths Chapter 15 Circumference and Area of a Circle Additional Questions 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 Additional Questions.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 15 |

Chapter Name |
Circumference and Area of a Circle |

Exercise |
Additional Questions |

Number of Questions Solved |
28 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 10 Maths Chapter 15 Circumference and Area of a Circle Additional Questions

**Multiple Choice Questions**

Question 1.

If circumference of a circle and perimeter (RBSESolutions.com) of a square are same then, [NCERT Exemplar Problem]

(A) Area of circle = Area of square

(B) Area of circle > Area of square

(C) Area of circle < Area of square

(D) Nothing is definite

Solution :

(B) is correct option.

Question 2.

If sum of area of the circles of radius R_{1} and R_{2} is equal to the area of circle of radius R, then [NCERT Exemplar Problem]

(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) is correct.

Question 3.

In fig. OACB is a quadrant of circle with center O and (RBSESolutions.com) radius 3.5 cm. If OD = 2 cm, then area of shaded part will be :

(A) 6.482 cm^{2}

(B) 5.485 cm^{2}

(C) 4.485 cm^{2}

(D) 3.485 cm^{2}

Solution :

Area of shaded part = \(\frac { 1 }{ 4 }\)π(R^{2} – r^{2}),

(where R = outer radius, r = Inner radius)

Thus, option (A) is correct.

Question 4.

Area of circle inscribed in a square of (RBSESolutions.com) side 6 cm will be : [NCERT Exemplar Problem]

(A) 36π cm^{2}

(B) 18π cm^{2}

(C) 12π cm^{2}

(D) 9π cm^{2}

Solution :

Given

Side of square ABCD = 6 cm

Radius of circle inscribe in a square of radius (r) = \(\frac { 6 }{ 2 }\) = 3 cm

Area of circle = πr^{2}

= π × (3)^{2}

= 9π cm

Thus (D) is correct.

Question 5.

The area of square inscribed in a (RBSESolutions.com) circle of radius 8 cm will be : [NCERT Exemplar Problem]

(A) 256 cm^{2}

(B) 128 cm^{2}

(C)64 \(\sqrt { 2 }\) cm^{2}

Solution :

Given

Radius of circle (r) = 8 cm

∴ Radius of circle = 2 × 8 = 16 cm

∵ We know that diagonals of square (RBSESolutions.com) inscribed in a circle bisects at center of circle.

Thus AC = 16 cm

Let side of square = x cm

In right angled triangle ABC

Thus option (B) is correct.

Question 6.

If perimeter of a circle is equal to the (RBSESolutions.com) perimeter of a square then ratio of their area will be : [NCERT Exemplar Problem]

(A) 22 : 7

(B) 14 : 11

(C) 7 : 22

(D) 11 : 14

Solution :

Let r is radius of circle and x is side of square.

Perimeter of circle = Perimeter of square

Area of circle : Area of square = 14 : 11

Thus, option (B) is correct.

Question 7.

The area of largest triangle which can be (RBSESolutions.com) drawn ¡n a semicircle of radius r : [NCERT Exemplar Problem]

(A) r^{2}

(B) \(\frac { 1 }{ 2 }\)r^{2}

(C) 2r^{2}

(D) \(\sqrt { 2 }\)r^{2}

Solution :

Let AB is diameter of semi-circle.

whose center is O and ΔABC is largest triangle which can be drawn in semicircle in which AC = BC

∠ACB = 90° (angle in semicircle is 90°)

Thus, ACB is right triangle

Let AC = BC = x

Radius of (RBSESolutions.com) semicircle = r (given)

In right angled triangle ABC,

Thus, option(A) is correct.

Question 8.

If π = \(\frac { 22 }{ 7 }\), then (RBSESolutions.com) distance covered in the revolution (in m) of a wheel of radius 35 cm will be : [CBSE 2013]

(A) 2.2

(B) 1.1

(C) 9.625

(D) 96.25

Solution :

Given

Diameter of wheel = 35 cm

∴ Radius of wheel (r) = \(\frac { 35 }{ 2 }\) cm

Distance covered in 1 revolution by wheel = circumference of wheel

= 2πr

= 2 × \(\frac { 22 }{ 7 }\) × \(\frac { 35 }{ 2 }\)

= 110 cm = 1.1 m

Thus, option (B) ¡s correct.

Question 9.

The radius of a circle of circumference (RBSESolutions.com) equal to the sum of circumference of two circles of diameter 36 cm and 20 cm will be : [NCERT Exemplar problem]

(A) 56 cm

(B) 42 cm

(C) 28 cm

(D) 16 cm

Solution :

Diameters of two circles are 36 cm and 20 cm respectively so their radius r_{1} = \(\frac { 36 }{ 2 }\) = 18 cm and r_{2} = \(\frac { 20 }{ 2 }\) = 10 cm

Let R be the radius of required circle.

According to the question

Sum of circumference of (RBSESolutions.com) given circles = Circumference of required circle

Thus, option (C) is correct.

Question 10.

The diameter of a circle of area equal to sum of areas of (RBSESolutions.com) two circles of radius 24 cm and 7 cm will be : [NCERT Exemplar Problem]

(A) 31 cm

(B) 25 cm

(C) 62 cm

(D) 50 cm

Solution :

Given :

Radius of two circles r_{1} = 24 cm, r_{2} = 7 cm

Let R be the radius of required circle.

According to question

Sum of areas of two given circles = area of required circle

Thus, diameter of required circle = 2 × 25 = 50 cm

Thus, (D) is correct.

**Short/Long Answer Type Questions**

Question 1.

If radius of a circle is 14 cm, then find (RBSESolutions.com) area of circle. [M.S.N. Raj 2013]

Solution :

Given

Radius of circle (r) = 14 cm

Area of circle = πr^{2}

= \(\frac { 22 }{ 7 }\) × (14)^{2}

= 22 × 14 × 2

= 616 sq cm

Thus, Area of circle = 616 sq cm

Question 2.

Find the area of sector of a circle of radius r, whose (RBSESolutions.com) angle is θ in degree. (MS.B. Raj 2014)

Solution :

Area of sector of circle whose radius r and angle is in degree θ, is

= \(\frac { { \pi r }^{ 2 }\theta }{ { 360 }^{ \circ } }\)

Question 3.

A sector of circle of radius r, whose angle In degree is O. Find the length of arc. (M.S.B. Raj 2015)

Solution :

A sector of circle of radius r, whose angle in degree is 9 then length of arc (l) = \(\frac { \pi r\theta }{ { 180 }^{ \circ } } \)

Question 4.

Find the area of circle whose (RBSESolutions.com) circumference is 44 cm. (M.S.B. Raj 2014)

Solution :

Given :

Circumference of circle = 44 cm

We know that

Circumference of circle = 2πr

44 = 2 × \(\frac { 22 }{ 7 }\) × r

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

Area of circle = πr^{2}

= \(\frac { 22 }{ 7 }\) × 7 × 7

= 154 sq cm

Question 5.

Find the area of major sector (RBSESolutions.com) corresponding to angle 120° In circle of radius 7 cm. (M.S.B. Raj 2015)

Solution :

Given : θ = 120° radius r = 7 cm

Area of major sector

Question 6.

Find the circumference of a (RBSESolutions.com) circle of diameter 14 cm. (M.S.B. 2015)

Solution :

Given

Diameter of circle = 14 cm

We know that

Radius r = \(\frac { 14 }{ 2 }\) = 7 cm

Circumference of circle = 2πr

= 2 × \(\frac { 22 }{ 7 }\) × 7

= 44 cm

Question 7.

The diameter of a circular pond is 17.5 m. It is enclosed by a circular path of width 2 m. Find the cost of making this path at ₹ 25/m^{2}. (NCERT Exemplar Problem)

Solution :

Let O is center of circular pond whose (RBSESolutions.com) diameter is 17.5 m and it is enclosed by a 2 m broad path. So radius of circular path.

r = \(\frac { 17.5 }{ 2 }\) m

Radius of pond along with path (R)

\(\frac { 17.5 }{ 2 }\) + 2

\(\frac { 21.5 }{ 2 }\) m

Area of circular path

Making cost of path = 122.46 × 25 = ₹ 3061.50

Thus, cost of path = ₹ 3061.50

Question 8.

An arc of circle of radius 21 cm, subtends an (RBSESolutions.com) angle 60° at center, then find area of corresponding major sector. (BSER 2012)

Solution:

Radius (r) = 21 cm

Let arc AB subtends angle 60° at center corresponding major sector angle

θ = 360°- 60° = 300°

Area of corresponding major sector

Thus area of corresponding major sector = 1155 cm^{2}

Question 9.

Find the area of sector of circle (RBSESolutions.com) of radius 4 cm, whose angle is 60°. Also find area of corresponding major sector [π = 3.14] [M.S.R. Raj 2013]

Solution:

Radius of circle (r) = 4 cm

Angle of sector r(θ) = 60°

Angle of corresponding major sector = 360° – 60° = 300°

∴ Area of corresponding major sector

Thus area of sector = 8.37 cm^{2} (approx)

area of major sector = 41.87 cm^{2} (approx)

Question 10.

ln given figure, find the area of shaded part If ABCD is a (RBSESolutions.com) square of side 14 cm and APD and BPC are two semicircles. (CBSE 2012)

Solution :

Given: Side of square = 14 cm

Diameter of semi circle (AD = CB) = 14 cm

Radius of (RBSESolutions.com) semicircle (R) = \(\frac { 14 }{ 2 }\) = 7 cm

∴ Area of square ABCD = (side)^{2}

= 14 × 14 = 196 cm^{2}

Area of semicircle = \(\frac { 1 }{ 2 }\)πr^{2}

\(\frac { 1 }{ 2 }\) × \(\frac { 22 }{ 7 }\) × 7 × 7 = 77 cm^{2}

Area of two semicircle =2 × 77 = 154 cm^{2}

∴ Area of shaded part Area of square ABCD – Area of two semicircles

= 196 – 154 = 42 cm^{2}

Thus area of shaded part 42 sq cm

Question 11.

From each corner of square of side 4 cm, quadrant of (RBSESolutions.com) circle of radius 1 cm is cut and also a circle of 2 cm diameter is cut from middle as shown in figure. Find the area of remaining part of square. (CBSE 2012)

Solution :

Let ABCD is a square.

Its each side = 4 cm

∴ Area of square = (side)^{2}

= (4)^{2} = 16 cm^{2}

Radius of quadrant of circle (r) = 1 cm

∴ Area of four (RBSESolutions.com) quadrants = 4 × \(\frac { 1 }{ 4 }\)πr^{2}

= πr^{2}

= \(\frac { 22 }{ 7 }\) × (1)^{2}

= \(\frac { 22 }{ 7 }\) sq cm

∵ Diameter of circle cut in center = 2 cm

Radius (R) = \(\frac { 2 }{ 2 }\) = 1 cm

∴ Area of circle = πr^{2}

= \(\frac { 22 }{ 7 }\) × (1)^{2}

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

Now area of shaded part = Area of square – area of 4 quadrants – area of circle

= 16 – \(\frac { 22 }{ 7 }\) – \(\frac { 22 }{ 7 }\)

= \(\frac { 112-22-22 }{ 7 }\) = \(\frac { 68 }{ 7 }\)

Thus, area of shaded part = \(\frac { 68 }{ 7 }\) sq cm

Question 12.

In the following figure, find the area of shaded (RBSESolutions.com) portion. If AB = 5 cm, AC = 12 cm and O is center of circle. (MS.B. Raj. 2014)

Solution :

From figure it is clear that ∠BAC is angle is semicircle.

Thus, this a (RBSESolutions.com) right angled triangle

By Pythagoras theorem,

BC^{2} = AB^{2} + AC^{2}

= (5)^{2} + (12)^{2}

= 25 + 144

= 169

BC = 13 cm

Radius of circle R = \(\frac { 13 }{ 2 }\)cm

∴ Area of shaded portion = Area of semicircle – Area of ΔABC

Thus, area of shaded portion = \(\frac { 1019 }{ 28 }\) sq.cm

Question 13.

Find the area of minor segment of (RBSESolutions.com) circle of radius 14 cm, whose central angle in 60°. Also find area of corresponding major segment. \(\left( \pi =\frac { 22 }{ 7 } \right) \) (CBSE 2015)

Solution :

Given : Radius of circle = 14 cm

Angle subtended chord AB at O is 60°

Now Area of circle = πr^{2}

= \(\frac { 22 }{ 7 }\) × (14)^{2} = 616 cm^{2}

Area of sector AOB

Area of minor (RBSESolutions.com) segment = A a of sector AOB – Area of ΔAOB

= 102.67 – 84.87

= 17.8 cm^{2} (approx)

Area of major segment = Area of circle – Area of minor segment

= 616 – 17.8

= 598.2 cm^{2}

Question 14.

In adjoining figure, OACB is a quadrant of a circle (RBSESolutions.com) with center O and radius 3.5 cm. If OD = 2 cm, then find the area of the following :

(i) Quadrant OACB

(ii) Shaded part

Solution :

Radius of circle (r) =3.5 cm and OD = 2 cm

(i) Area of quadrant OACB

= \(\frac { 1 }{ 4 }\) × πr^{2}

= \(\frac { 1 }{ 4 }\) × \(\frac { 22 }{ 7 }\) × (3.5)^{2}

= \(\frac { 38.5 }{ 4 }\) cm^{2}

= \(\frac { 77 }{ 8 }\) cm^{2} = 9.625 cm^{2}

(ii) Area of quadrant OACB – Area of right triangle ΔOBD

= \(\frac { 77-28 }{ 8 }\) = \(\frac { 49 }{ 8 }\) cm^{2}= 6.125 cm^{2}

Thus, area of shaded part = 6.125 cm^{2}

Question 15.

In given fig, O is center of the circle in (RBSESolutions.com) which AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of shaded part. [π = 3.14] [CBSE – 2012]

Solution :

Given : O is center of circle. So BC is diameter of circle.

∠CAB = 90° (∵ Angle in semicircle is 90°)

∴ CAB is a right angled triangle. in which AC = 24 cm and AB = 7 cm

In right angled triangle CAB

BC^{2} = AC^{2} + AB^{2} (By Pythagoras Theorem)

⇒ BC^{2} = 24^{2} + 7^{2}

⇒ BC^{2} = 576 + 49

⇒ BC^{2} = 625

⇒ BC = \(\sqrt { 625 }\) = 25 cm

⇒ Radius (OC) = \(\frac { 25 }{ 2 }\) cm

Now, area of shaded part = Area of (RBSESolutions.com) semicircle – area of right angled ΔCBD + area of quadrant BOD

So, area of shaded part = 283.97 cm^{2} (approx)

Question 16.

The sides of a triangles field are 15 m, 16 m and 17 m. For grazing in (RBSESolutions.com) field a cow a buffalo and a horse are tied at each corner of field by 7 m long rope separately. Find the area of that portion of field which will not be grazed by these animals. [NCERT Exemplar Problem]

Solution :

Let ABC ¡s triangular field with sides AB = 15 cm, BC = 16 m and AC = 17 m. A cow, a buffalo and a horse are tied at vertices of field A, B and C respectively and radius of each sector is 7 cm.

Question 17.

Three circles of equal radius 3.5 cm are drawn In which (RBSESolutions.com) each touches other two find the area enclosed by these circles. (NCERT Exemplar Problem)

Solution :

Taking A, B, and C as centers and radius 3.5 cm draw three circles in which each touches other two. Join AB, BC and CA.

AB = 3.5 + 3.5 = 7 cm

BC = 3.5 + 3.5 = 7 cm

AC = 3.5 + 3.5 = 7 cm

Thus, ABC ¡s an equilateral triangle

∠A = ∠B = ∠C = 60°

Area of sector with center A

∵ Radius and angle of three (RBSESolutions.com) sectors are same

∴ Area of three sectors = \(\frac { 3\times 38.5 }{ 6 } \)

= 19.25 cm^{2}

Area of equilateral ΔABC

= \(\frac { \sqrt { 3 } }{ 4 } \) × (side)^{2}

= \(\frac { 1.732 }{ 4 }\) × (7)^{2} = 21.217 cm^{2}

Area of enclosed by three circles = Area of shaded part = Area of equilateral triangle – Area of three sectors

= 21.217 – 19.25

= 1.967 cm^{2} (approx)

Thus, area enclosed by three cicrles = 1.967 cm^{2} (approx)

Question 18.

In gives figure, in a quadrant OPBQ of a circle, (RBSESolutions.com) a square OABC is inscribed. If OA = 1 cm, then find area of shaded part.\(\left( \pi =\frac { 22 }{ 7 } \right) \)

Solution :

∵ OABC is a square

∴ OA = AB = BC = OC = 21 cm

In right angled ΔOAB

OB^{2} = OA^{2} + AB^{2}

= (21)^{2} + (21)^{2}

= 441 + 441 = 882

⇒ OB = \(\sqrt { 882 }\) = 21\(\sqrt { 2 }\) cm

∴ Radius of (RBSESolutions.com) quadrant (r) = 21\(\sqrt { 2 }\) cm

Angle of sector (θ) = 90°

Area of quadrant OPBQ

Area of square OABC = 21 × 21 = 441 cm^{2}

∴ Area of shaded part

= Area of quadrant OPBQ – Area of square OABC

= 693 – 441 = 252 cm^{2}

Thus area of shaded part = 252 cm^{2}

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