RBSE Solutions for Class 7 Maths Chapter 16 Perimeter and Area Ex 16.3 is part of RBSE Solutions for Class 7 Maths. Here we have given Rajasthan Board RBSE Class 7 Maths Chapter 16 Perimeter and Area Exercise 16.3.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 7 |

Subject |
Maths |

Chapter |
Chapter 16 |

Chapter Name |
Perimeter and Area |

Exercise |
Ex 16.3 |

Number of Questions |
16 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 7 Maths Chapter 16 Perimeter and Area Ex 16.3

Question 1

Find the circumference of circle having (RBSESolutions.com) following radius (π = [latex]\frac { 22 }{ 7 }[/latex])

(i) 21 cm

(ii) 28 cm

(iii) 10.5 cm

Solution:

(i) Radius of circle = 21 cm

Circumference of Circle = 2πr

= 2 x [latex]\frac { 22 }{ 7 }[/latex] x 21 = 2 x 22 x 3 = 44 x 3

= 132 cm

(ii) Radius of circle = 28 mm

circumference = 2πr

= 2 x [latex]\frac { 22 }{ 7 }[/latex] x 28 = 2 x 22 x 4 = 2 x 88

= 176cm

(iii) Radius of circle = 10.5 cm

Circumference of circle = 2πr

= 2 x [latex]\frac { 22 }{ 7 }[/latex] x 10.5

= 2 x 22 x 1.5

= 44 x 1.5 = 66 cm.

Question 2

Find the area of the (RBSESolutions.com) following circles. Given

(i) Radius = 5 cm

(ii) Diameter = 42 metre

(iii) Radius = 5.6 cm

Solution:

(i) Radius (r)= 5 cm

Area of circle = πr^{2}

= 3.14 x (5)^{2} = 3.14 x 25 = 78.57 sq. cm

(ii) Diameter (d) = 42 m

Radius (r) = [latex]\frac { 42 }{ 2 }[/latex] = 21 m

Area of circle = πr^{2}

= [latex]\frac { 22 }{ 7 }[/latex] x 21 x 21 = 22 x 3 x 21

= 66 x 21

= 1386 sq. cm

(iii) Radius (r) = 5.6 cm

Area of circle = πr^{2}

= [latex]\frac { 22 }{ 7 }[/latex] x 5.6 x 5.6 = 22 x 0.8 x 5.6 = 98.56 sq. cm.

Question 3

Find the radius of a circular sheet (RBSESolutions.com) whose perimeter is 132 metre. Also find its area (π = [latex]\frac { 22 }{ 7 }[/latex])

Solution:

Circumference of circle = 132 m

Let the radius of circle is r metre

Circumference = 132 ⇒ 2πr = 132

⇒ 2 x [latex]\frac { 22 }{ 7 }[/latex] x r = 132

⇒ r = [latex]\frac { 132\times 7 }{ 22\times 2 }[/latex] = 21 cm

Area of sheet = πr^{2}

= [latex]\frac { 22 }{ 7 }[/latex] x 21 x 21 = 22 x 3 x 21

= 66 x 21

= 1386 sq. m

Question 4

Circumference of a circle is 44 cm. Find the (RBSESolutions.com) radius and area of triangle. (π = [latex]\frac { 22 }{ 7 }[/latex])

Solution:

Let radius of cirlce r cm

Circumference of circle = 44

⇒ 2πr = 44

⇒ 2 x [latex]\frac { 22 }{ 7 }[/latex] x r = 44

r = [latex]\frac { 44\times 7 }{ 22\times 7 }[/latex] = 7 cm

∴ Area of circle = πr^{2}

= [latex]\frac { 22 }{ 7 }[/latex] x 7 x 7 = 22 x 7 = 154sq. cm

Question 5

The given figure is a semi circle with 12 cm. diameter. Find out (RBSESolutions.com) its circumference.

Solution:

Diameter (d)

Radius r = [latex]\frac { 12 }{ 2 }[/latex] = 6 cm

Perimeter of half circle = πr + diameter

= 3.14 x 6 + 12 (∵ π = 3.14)

= 30.84 cm.

Question 6

The radius of a circular pond is 28 meter. 4 path of 1.4 meter width is (RBSESolutions.com) present around it Find the area of path.

Solution:

Radius of circular pond (r) = 28 m

Radius of circular pond including path

(R) = 28 + 1.4

= 29.4 m

Area of path

= πR^{2} – 4^{2}

= π(R^{2} – r^{2})

= π(29.4^{2} – 28^{2})

= [latex]\frac { 22 }{ 7 }[/latex](864.36 – 784)

= [latex]\frac { 22 }{ 7 }[/latex] x 80.36

= 252.56 sq. m

Question 7

Area of a circle is 616 square cm. The circle is (RBSESolutions.com) surrounded by a path 2 meter wide. What is the area of this path?

Solution:

Area of path = 616 sq. cm

Let radius of circle is r cm

Area of circle = 616 sq. cm

⇒ πr^{2}= 616

⇒ [latex]\frac { 22 }{ 7 }[/latex] x r^{2} = 616

⇒ r^{2} = [latex]\frac { 616\times 7 }{ 22 }[/latex]

⇒ r^{2} = 28 x 7

⇒ r^{2} = 196

⇒ r = 14 cm

After 2 cm wide road let the radius is

R = r + 2

R = 14 + 2 = 16 cm

Area of path = = πR^{2} – πr^{2}

= π(R^{2} – r^{2}) = [latex]\frac { 22 }{ 7 }[/latex] x (16^{2} – 14^{2})

= [latex]\frac { 22 }{ 7 }[/latex] x (256 – 196) = [latex]\frac { 22 }{ 7 }[/latex] x 60

= 188.57 sq. cm.

Question 8

From a circular card sheet of radius 5 cm, a circular sheet (RBSESolutions.com) of radius of 4 cm is removed. Find the area of the remaining sheet. (π = 3.14)

Solution:

Area of sheet (R) = 5 cm

Radius of circle (r) = 4 cm

Area of remaining part = = πR^{2} – πr^{2}

= π(R^{2} – r^{2})

= 3.14(5^{2} – 4^{2})

= 3.14(25 – 16)

= 3.14 x 9

= 28.26 sq. cm

Question 9

From a circular card sheet of radius 14 cm, a square of 4 cm is (RBSESolutions.com) removed as shown in the adjoining figure. Find the area of the remaining sheet

(π = [latex]\frac { 22 }{ 7 }[/latex])

Solution:

Area of remining sheet = Area of sheet – Area of square

= πr^{2} – 4^{2} = [latex]\frac { 22 }{ 7 }[/latex] x 14 x 14 – 16

[∵ r = 14 cm]

= 22 x 2 x 14 – 16

= 616 – 16

= 600 sq. cm

Question 10

The ratio of diameter of two circles is 4 : 5 then (RBSESolutions.com) Find out ratio of their areas.

Solution:

Let the diameter are : d_{1} and d_{2}

d_{1}: d = 4 : 5 or [latex]\frac { { d }_{ 1 } }{ { d }_{ 2 } }[/latex] = [latex]\frac { 4 }{ 5 }[/latex]

Let the radius of circles are r_{1} and r_{2} and their areas are A_{1} AND A_{2} then,

r_{1} = [latex]\frac { { d }_{ 1 } }{ 2 }[/latex], r = [latex]\frac { { d }_{ 2 } }{ 2 }[/latex]

Question 11

Durga want to polish a circular table top of diameter 2.8 metre. Find (RBSESolutions.com) the cost of polishing if the rate of polishing is ₹25 per square meter.

Solution:

Diameter of table = 2.8 m

Radius of table (r) = [latex]\frac { 2.8 }{ 7 }[/latex] = 1.4 m

Area of table = πr^{2}

= [latex]\frac { 22 }{ 7 }[/latex] x 1.4 x 1.4 = 22 x 0.2 x 1.4

= 4.4 x 1.4 = 6.16 sq. m

cost of polishing per sq.m = ₹25

6.16 वर्ग मीटर टेबल पर पॉलिश कराने का खर्चा = 25 x 6.16 = ₹154

Question 12

Gopi ties his horse with a rope of length 12 m. What (RBSESolutions.com) area of grass can the horse graze upon ?

Solution:

Radius of rope (r) = 12 m

Area of land grazed by horse = πr^{2}

= 3.14 x (12)^{2} = 3.14 x 144

= 452.16 sq.m

Question 13

In the given figure, two semicircular parts of diameter 12 cm, are add on (RBSESolutions.com) both the ends of a rectangular part Length of the part is 15 cm. Find out the area ?

Solution:

Diameter of table = 12 cm

Radius (r) = [latex]\frac { 12 }{ 2 }[/latex] = 6 cm

Area of ABCD

= length x breadth = 15 x 12 = 180 sq. cm

Area of two semi circles = 2 x [latex]\frac { 1 }{ 2 }[/latex] πr^{2}

= 3.14 x 6 x 6 = 3.14 x 36 = 113.04sq. cm

Area of whole shape = Area of rectangle + Area of two semi circle

= 180+113.04 = 293.04 sq. cm.

Question 14

How many rounds does a wheel of radius 35 take in order to cover (RBSESolutions.com) a distance of 880 metre? (π = [latex]\frac { 22 }{ 7 }[/latex])

Solution:

Radius of wheel (r) = 35 m

Circumference of wheel = 2πr

= 2 x [latex]\frac { 22 }{ 7 }[/latex] x 35 = 2 x 22 x 5 = 44 x 5

= 220 m

Wheel covers 220 metre distance revolution.

So to cover a distance of 880 metre number of revolution = [latex]\frac { 880 }{ 220 }[/latex] = 4 revolution.

Question 15

Parvat has spend how much money to pour the soilalla (RBSESolutions.com) round it circular shape garden in 7 m wide road. It the diameter of garden is 56 metre? (π = [latex]\frac { 22 }{ 7 }[/latex])

Solution:

Diameter of park = 56 m 56

Radius (r) = [latex]\frac { 56 }{ 2 }[/latex] = 28 m

The radius of (RBSESolutions.com) park including path (R) = 28 + 7

= 35 m

Area of path = = πR^{2} – πr^{2}

= π(35^{2} – 28^{2}) = [latex]\frac { 22 }{ 7 }[/latex] x (35 + 28) (35 – 28)

= [latex]\frac { 22 }{ 7 }[/latex] x 63 x 7 = 22 x 63 = 1386 sq.m

Expenditure of putting soil = Area of road x Expenditure per sq. m

= 1386 x 11 = ₹15,246

Question 16

Length of minute arm of circular shape clock is 20 cm. How (RBSESolutions.com) much distance does tip of minute hand cover in one hour = π = 3.14.

Solution:

Length of minute hand = 20 cm

r = 20 cm

So the distance covered by minute hand is one completer revolution

= 2πr = 2 x 3.14 x 20 = 125.6 cm

We hope the RBSE Solutions for Class 7 Maths Chapter 16 Perimeter and Area Ex 16.3 will help you. If you have any query regarding Rajasthan Board RBSE Class 7 Maths Chapter 16 Perimeter and Area Exercise 16.3, drop a comment below and we will get back to you at the earliest.

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