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RBSE Solutions for Class 5 Maths Chapter 14 Perimeter and Area In Text Exercise

March 13, 2019 by Fazal Leave a Comment

   

RBSE Solutions for Class 5 Maths Chapter 14 Perimeter and Area In Text Exercise is part of RBSE Solutions for Class 5 Maths. Here we have given Rajasthan Board RBSE Class 5 Maths Chapter 14 Perimeter and Area In Text Exercise.

Board RBSE
Textbook SIERT, Rajasthan
Class Class 5
Subject Maths
Chapter Chapter 14
Chapter Name Perimeter and Area
Exercise In Text Exercise
Number of Questions 4
Category RBSE Solutions

Rajasthan Board RBSE Class 5 Maths Chapter 14 Perimeter and Area In Text Exercise

Page No. 72
Question 1.
Nirmal’s mathematics book has questions to find the perimeter of some shapes :
RBSE Solutions for Class 5 Maths Chapter 14 Perimeter and Area In Text Exercise image 1
Can you find out the perimeter of above three shapes?
Solution.
We know that, only closed shapes have perimeters. Here(RBSESolutions.com)shape (b) and (c) are not closed, so we cannot find their perimeter. These type of shapes are known as open shapes.

RBSE Solutions

Page No. 72
Question 1.
Some shapes are given below :
RBSE Solutions for Class 5 Maths Chapter 14 Perimeter and Area In Text Exercise image 2
RBSE Solutions for Class 5 Maths Chapter 14 Perimeter and Area In Text Exercise image 3
Find out the perimeter of these shapes.
Solution.
(a) We know that Perimeter(RBSESolutions.com)of regular shapes = number of sides × length of a side
∴ Perimeter = 3 × 10 = 30 cm.
(b) Perimeter = 4 × 12 = 48 cm.
(c) Perimeter = 5 × 7 = 35 cm.

RBSE Solutions

Page No. 73
Try these
Question 1.
RBSE Solutions for Class 5 Maths Chapter 14 Perimeter and Area In Text Exercise image 4
Think, is rectangle a regular shape? If you want to find a formula for the perimeter of a rectangle. What will you do?
Solution.
No, rectangular is not a regular shape.
Because it’s all sides are not equal measurement.
We know that opposite sides of a rectangle are equal.
i.e. perimeter = 2 × length + 2 × breadth
Therefore formula(RBSESolutions.com)for the perimeter of rectangle
perimeter = 2 × (length + breadth)
This is the perimeter of rectangle.
Now perimeter of given rectangular shape = 2 × (15 + 5)
= 2 × 20 = 40 m.

RBSE Solutions

Page No. 75
Question 1.
Some shapes are given below on grid paper. Find the area of these.
RBSE Solutions for Class 5 Maths Chapter 14 Perimeter and Area In Text Exercise image 5
Solution.
(i) Shape is a picture of rectangle.
If each side of square is of 1 cm on grid paper then
length of rectangle is = 5 cm
and width = 2 cm.
Number of squares covered by rectangle ABCD = 10
Therefore, area of(RBSESolutions.com)rectangle ABCD = 10 square cm.
Here, there is a rule been between length, width and area of rectangle
Length × Width = Area
= 5 cm. × 2 cm. = 10 square cm.

RBSE Solutions

(ii) Length of square = 4 cm.
Width of square = 4 cm.
Area of square = (side)2 = (4)2 = 16 square cm.
Number of squares covered by square = 16.
Therefore area of square = 16 square cm.

(iii) Length of rectangle = 7 cm.
Width of rectangle = 3 cm.
Area of rectangle = Length × Width = 7 × 3 = 21 square cm.
Number of(RBSESolutions.com)squares covered by rectangle ABCD = 21.
Therefore area = 21 square cm.

RBSE Solutions

We hope the RBSE Solutions for Class 5 Maths Chapter 14 Perimeter and Area In Text Exercise will help you. If you have any query regarding Rajasthan Board RBSE Class 5 Maths Chapter 14 Perimeter and Area In Text Exercise, drop a comment below and we will get back to you at the earliest.

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