RBSE Solutions for Class 7 Maths Chapter 16 Perimeter and Area Ex 16.1

RBSE Solutions for Class 7 Maths Chapter 16 Perimeter and Area Ex 16.1 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 16.1.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 7
Subject	Maths
Chapter	Chapter 16
Chapter Name	Perimeter and Area
Exercise	Ex 16.1
Number of Questions	7
Category	RBSE Solutions

Rajasthan Board RBSE Class 7 Maths Chapter 16 Perimeter and Area Ex 16.1

Question 1
Radha takes 2 round daily along the sides of a square (RBSESolutions.com) park of side 60 meter. Then find out that how much distance is covered by her daily?
Solution:
Side of square park = 60 m
Perimeter of square park = 4 x side
= 4 x 60 = 240 m
Distance covered by Radha everyday
= No. of rounds x Perimeter of park
= 2 x 240 = 480 m

Question 2
Suresh has a ribbon of length 78 cm. He wants to (RBSESolutions.com) apply it on the edges of a rectangular photo frame of length 26 cm. Find the breadth of frame.
Solution:
Length of photo frame = 26cm, Breadth ?
Length of ribbon Perimeter of photo frame
⇒ 78 = 2 (length + breadth)
⇒ 2 (length + breadth) = 78
⇒ 2(26 + breadth) 78
⇒ 26 + breadth = \(\frac { 78 }{ 2 }\)
⇒ 26 + breadth = 39
⇒ breadth = 39 – 26 = 13cm

Question 3
Ranu wants to spread a down the carpet in meeting hail whose (RBSESolutions.com) length is 50 meter if breadth is half of the length then find the area of carpet.
Solution:
Length of carpet = 50 m

Breadth of carpet = \(\frac { 1 }{ 2 }\) x length
= \(\frac { 1 }{ 2 }\) x 50 = 25 m
Area of carpet length x breadth
= 50 x 25 = 1250 m²

Question 4
Gurmeet sowed the crop of moong in 4260 square meter part of his land. He (RBSESolutions.com) wants to fence around the field. 1f the breadth of the land is 30 metre then how much wire (in length) is needed for it ?
Solution:
Area of field 4200 sq. m²
Breadth of field = 30 m
Length of field


Required wire for fencing = Perimeter of field 2 (length x breadth)
= 2(140 + 30) = 2(170) = 2 x 170 = 340m.

Question 5
Area of playground of school is 38400 square meter. If the ratio of (RBSESolutions.com) length and breadth of playground is 3 : 2 then find the perimeter of playground.
Solution:
Area of playground = 38400 sq. m
Let the length = 3x and breadth is 2x meter
Area of playground = length x breadth

⇒ 38400 = 3x × 2x
⇒ 38400 = 6x²
⇒ 6x²= 38400
⇒ x² = \(\frac { 384000 }{ 6 }\)
⇒ x² = 6400
⇒ x = \(\sqrt { 6400 }\) = 80
Length of playground = 3x = 3 x 80 = 240 m
Breadth of playground = 2x = 2 x 80 = 160m
Perimeter of playground = 2(l+ b)
= 2(240 + 160) = 2(400)
= 2 x 400
= 800 m

Question 6
The perimeter of a rectangle and a square are same, the (RBSESolutions.com) length and breadth of rectangle is 25 cm and 15 cm respectively. Which figure has greater area?
Solution:
Perimeter of rectangle = 2 (length x breadth)
= 2(25 + 15) = 2(40) = 2 x 40 = 80 cm
Perimeter of square = Perimeter of rectangle
4 x side = 80 ⇒ side = \(\frac { 80 }{ 4 }\) = 20 cm 4
⇒ Area of square = side² = 20 x 20 = 400 sq. cm
⇒ Area of rectangle = length x breadth
= 25 x 15 = 375 sq. cm
Area of square is 400 sq. cm and area of rectangle is 375 sq. cm.
Hence area of square has greater area.

Question 7
Find the perimeter (RBSESolutions.com) of following figures :
(i) Triangle whose sides are 2 cm, 3 cm and 4 cm.
(ii) Equilateral triangle with side 8 cm.
(iii) Isosceles triangle whose equal sides are 10 cm and the third side is 7 cm.
Solution:
(i) Sides of triangle: 2 cm, 3 cm and 4 cm
Perimeter of triangle = Sum of all sides
= 2 + 3 + 4 = 9 cm

(ii) Side of equilateral angle = 8 cm Perimeter (RBSESolutions.com) of equilateral triangle = 3 x side
= 3 x 8
= 24 cm

(iii) Two sides of isosceles triangle = 10 cm, 10 cm Third side = 7 cm
Perimeter of isosceles triangle
= 10 + 10 + 7
= 27 cm

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