AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 1

AP State Syllabus AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 1 Textbook Questions and Answers.

AP State Syllabus 7th Class Maths Solutions 12th Lesson Area and Perimeter Exercise 1

AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 1

Question 1.
AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 1 1
Solution:
AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 1 2

Question 2.
The measurements of some squares are given in the table below. However, they are incomplete. Find the missing information.
AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 1 3
Solution:
i) Perimeter = 4a = 4(15) = 60cm
ii) Perimeter = 4a = 88
a = \(\frac{88}{4}\)= 22
Side = 22cm
Area = a2 = (22)2 = 484 cm2

AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 1

Question 3.
The measurements of some rectangles are given in the table below. However, they are incomplete. Find the missing information.
AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 1 4
Solution:
Given:
i) Length = 20 cm
breadth = 14cm
Area = l x b = 20 x 14 =280 cm2
Perimeter = 2(l + b) = 2(20 + 14)
= 2(34) = 68cm

ii) Given breadth = 12cm, perimeter = 60cm
Perimeter 2(l + b) 60 cm
= 2(l + 12) =60cm
2l + 24 =60cm
2l = 60 – 24 = 36
2l = \(\frac{36}{2}\) = 18cm
Area = l x b = 18 x 12 = 216cm2

AP Board 7th Class Maths Solutions Chapter 13 Area and Perimeter Ex 1

iii) Given length = 15cm, Area = 150 cm
Area = l x b = 15 x b = 150
⇒ b = \(\frac{150}{15}\) = 10cm
Perimeter = 2(l + b) = 2(15 + 10)
= 2(25) =50cm

Leave a Comment