If you are preparing for the final board exam or any competitive exam, all you need to do is practice. For your benefit, our math subject experts have meticulously prepared a list of important questions for CBSE class 10 Math chapter-wise and topic-wise.
Important questions based on NCERT syllabus for Class 10 Chapter 12 - Areas Related to Circles:
Question 1: In the figure, ABCD is a square of side 14 cm with centres A, B, C, and D four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.
All the four circles are congruent.
Therefore, unshaded area of the square is equal to one full circle with radius 7 cm.
Area of square
= 142 = 196 cm^2
Area of circle
=π(7)2 = 153.86 cm^2
Area of shaded portion
= 196 − 153.86
= 42.14 cm^2
Question 2: The cost of fencing a circular field at the rate of Rs 24 per metre is Rs 5280. The field is to be ploughed at the rate of Rs 0.50 per m^2. Find the cost of ploughing the field in Rupees. (Take π = 22/7).
Length of the fence (in meters)
= Total cost/Rate = 5280/24 = 220
So, circumference of the field = 220 m
Therefore, if r meters is the radius of the field, then
2πr = 220
2 × 22/7 × r = 220
r = 220 × 7/22 × 2 = 35
i.e., radius of the field is 35 m.
Therefore, area of the field
= πr^2 = 22/7 × 35 × 35 = 22 × 5 × 35 m^2
Now, cost of ploughing 1 m2 of the field = Rs. 0.50
So, total cost of ploughing the field
= Rs. 22 × 5 × 35 × 0.50 = Rs.1925
Question 3: In Akshita’s house there is a flower pot. The sum of radii of circular top and bottom of a flowerpot is 140 cm and the difference of their circumference is 88 cm. Find the diameter of the circular top and bottom.
Diameter on top = R
Radius of bottom = r
R + r = 140 cm….(1)
2πR − 2πr = 88cm
R − r = 88/2π = 14 ….(2)
Adding (1) and (2),
2R = 154
R = 77
r = 140 − 77 = 63
Thus diameter of top will be 154cm and diameter of bottom is 126cm