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

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 7 |

Subject |
Maths |

Chapter |
Chapter 16 |

Chapter Name |
Perimeter and Area |

Exercise |
Additional Questions |

Number of Questions |
16 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 7 Maths Chapter 16 Perimeter and Area Additional Questions

**Multiple Choice Questions**

Question 1

Formula to find out (RBSESolutions.com) area of circle is :

(A) πr^{2}

(B) 2πr^{2}

(C) 2πr

(D) πr

Question 2

The radius of circular shape field is 14 cm then circumference of field will be :

(A) 48 cm

(B) 88 cm

(C) 68 cm

(D) 98 cm

Question 3

Formula of area (RBSESolutions.com) of rectangle is :

(A) l x b

(B) l^{2}

(C) l x b^{2}

(D) b^{2}

Question 4

The area of 10 cm side square will be :

(A) 101 sq. m

(B) 100 sq. cm

(C) 100 sq. m

(D) 100 km

Question 5

The base of a triangle is 12 cm and height is 5 cm (RBSESolutions.com) then area of triangle will be :

(A) 30 sq. cm

(B) 60 sq. cm

(C) 120 sq. cm

(D) 28 sq. cm

Question 6

The formula of the area of right angle tri¬angle will be :

(A) \(\frac { 1 }{ 2 }\) x Base x Height

(B) Base x Height

(C) 2 x Base x Height

(D) 2 x Base + Height

Answer

1. (A), 2. (B), 3. (A), 4. (B), 5. (A), 6. (A).

**True/False**

(i) Half circumference (RBSESolutions.com) of circle = πr.

(ii) Area of parallelogram = \(\frac { 1 }{ 2 }\) x Base x Height

(iii) 1 km = 1000 m

(iv) π = \(\frac { 22 }{ 7 }\)

Ans

(i) True,

(ii) False,

(iii) True,

(iv) True.

**Fill in the blanks :**

(i) The perimeter of rectangle = ………….

(ii) Area of square = ……………

(iii) 1 m = ……….. cm

Answer

(i) 2(length + breadth) unit,

(ii) side^{2}

(iii) 100

**Very Short Answers Type Questions**

Question 1

Find out the circumference of circle (RBSESolutions.com) whose radius is 21 cm. (π = \(\frac { 22 }{ 7 }\) )

Solution:

Radius of circle (r) = 21 cm

Circumference of circle = 2πr

= 2 x \(\frac { 22 }{ 7 }\) x 21= 2 x 22 x 3 = 2 x 66 = 132 cm

Question 2

Find the area of a circle with radius 10 cm.

Solution:

Radius (RBSESolutions.com) of circle (r) = 10 cm

Area of circle = πr^{2}

= 3.14 x (10)^{2}

= 3.14 x 100

= 314 sq. cm.

Question 3

A rectangle’s length is 10 m and (RBSESolutions.com) breadth 5 m then find its perimeter.

Solution:

Perimeter of rectangle = 2 (length + breadth)

= 2(10 + 5) = 2(15) = 30 m

Question 4

A square whose sides is 12 m. Find its perimeter.

Solution:

Side of square = 12 m

Perimeter of square = 4 x side

= 4 x 12 = 48 m.

**Long Answer Questions**

Question 1

The area of rectangular sheet is 500 cm^{2}. If length (RBSESolutions.com) of sheet is 25 cm then what will be its breadth. Also find out perimeter of the sheet.

Solution:

Area of rectangular sheet = 500 cm

Length (l) = 25 cm

Area of rectangle sheet = l x b

(where b = breadth of sheet)

∴ breadth (b) = \(\frac { Area }{ Length }\) = \(\frac { 500 }{ 25 }\)

= 20 cm

Perimeter of (RBSESolutions.com) rectangular sheet

= 2 x (l + b)

= 2 x (25 + 20) cm

= 90 cm

So breadth of sheet is 20 cm and its perimeter is 90 cm.

Question 2

Find the perimeter of given shape.

Solution:

In this shape we have to find out perimeter, All (RBSESolutions.com) around of a sqaure, semi circles are joined.

The diameter of every semi circle is 14 cm

Circumference of a cirlce = πd (∴2r = d)

Half circumference = \(\frac { 1 }{ 2 }\)πd

= ( \(\frac { 1 }{ 2 }\) x \(\frac { 22 }{ 7 }\) x 14) = 22cm

So, the perimeter of given shape = 4 x 22 = 88 cm.

Question 3

Sudhanshu divides a 7 cm radius plate in two equal parts (RBSESolutions.com) find out the perimeter of every half plate take (π = \(\frac { 22 }{ 7 }\))

Solution:

Radius (r) = 7 cm

Circumference = 2πr

∴ Circumference of (RBSESolutions.com) half plate = \(\frac { 1 }{ 2 }\) x 2πr = πr

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

Diameter = 2r

= (2 x 7) cm = 14 cm

Perimeter of every half plate

= 22 cm + 14 cm = 36 cm.

Question 4

In a 100 side square park has 5 m wide path inside it bound. Find (RBSESolutions.com) out the area of path. Calculate the expenditure @ ₹250/- per 10m^{2} of cementing it.

Solution:

Let ABCD is 100 m side square park and shaded region is 5 m wide park.

PQ = 100 -(5+5) = 90m

Area of ABCD = (side)^{2}

= (100)^{2} sq. m = 10000 sq. m

Area of PQRS = (side)^{2}

= (90)^{2} sq. m

= 8100 sq. m

∴ Area of path = (10000 – 8100) sq. m

= 1900 sq. m

10 m^{2} cementing = ₹250

∴ Expense on 1900 m^{2} cementing 250

= \(\frac { 250 }{ 10 }\) x 1900

= ₹47,500

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

## Leave a Reply