Selina Concise Mathematics Class 9 ICSE Solutions Area and Perimeter of Plane Figures
ICSE SolutionsSelina ICSE Solutions
APlusTopper.com provides step by step solutions for Selina Concise Mathematics Class 9 ICSE Solutions Chapter 20 Area and Perimeter of Plane Figures. You can download the Selina Concise Mathematics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Mathematics for Class 9 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.
Download Formulae Handbook For ICSE Class 9 and 10
Selina ICSE Solutions for Class 9 Maths Chapter 20 Area and Perimeter of Plane Figures
Exercise 20(A)
Solution 1:

Solution 2:

Solution 3:

Solution 4:

Solution 5:
Since the perimeter of the isosceles triangle is 36cm and base is 16cm.


Solution 6:

Solution 7:

Solution 8:

Solution 9:

Solution 10:



Solution 11:

Solution 12:

Solution 13:

Solution 14:

Solution 15:

Exercise 20(B)
Solution 1:

Solution 2:

Solution 3:


Solution 4:


Solution 5:

Solution 6:

Solution 7:

Solution 8:

Solution 9:

Solution 10:

Solution 11:

Solution 12:

Solution 13:

We need to find the cost of carpeting of 80 cm = 0.8 m wide carpet, if the rate of carpeting is Rs. 25. Per metre.
Then

Solution 14:

Solution 15:

Solution 16:

Solution 17:

Solution 18:

Solution 19:

Solution 20:

Solution 21:

Solution 22:

Solution 23:

Solution 24:


Solution 25:

Solution 26:

Solution 27:



In the given figure, we can observe that the non-parallel sides are equal and hence it is an isosceles trapezium.

Solution 28:

Solution 29:

Solution 30:

Solution 31:

Solution 32:

Solution 33:

Solution 34:

Solution 35:

Solution 36:

Exercise 20(C)
Solution 1:

Solution 2:

Solution 3:

Solution 4:

Solution 5:

Solution 6:

Solution 7:

Solution 8:

Solution 9:

Solution 10:

Solution 11:

Solution 12:

Solution 13:

Solution 14:

Solution 15:

Solution 16:

Solution 17:

Solution 18:

Solution 19:

Solution 20:
From the given data, we can calculate the area of the outer circle and then the area of inner circle and hence the width of the shaded portion.

Solution 21:

Solution 22:

Solution 23:

Solution 24:

Solution 25:

Solution 26:

Solution 27:

Solution 28:

More Resources for Selina Concise Class 9 ICSE Solutions
- Selina Publishers Mathematics for Class 9
- Selina Class 9 ICSE Physics Solutions
- Selina ICSE Class 9 Chemistry Solutions
- Selina Concise Biology Class 9 Solutions