In the verge of coronavirus pandemic, we are providing FREE access to our entire Online Curriculum to ensure Learning Doesn't STOP!

Ex.11.3 Q10 Perimeter and Area - NCERT Maths Class 7

Go back to  'Ex.11.3'

Question

From a circular card sheet of radius \(14 \rm\,cm\), two circles of radius \(3.5\rm\, cm\) and a rectangle of length \(3\rm\, cm\) and breadth \(1 \rm\,cm\) are removed. (as shown in the adjoining figure). Find the area of the remaining sheet. (Take \(π =22/7\) )

 Video Solution
Perimeter And Area
Ex 11.3 | Question 10

Text Solution

What is known?

From a circular card sheet of radius \(14 \rm\,cm\), two circles of radius \(3.5\rm\, cm\) and a rectangle of length \(3\rm\, cm\) and breadth \(1 \rm\,cm\) are removed.

What is unknown?

The area of the remaining sheet.

Reasoning:

From a circular card sheet of radius \(14 \rm\,cm\),two circles of radius \(3.5\rm\, cm\) and a rectangle of length \(3\rm\, cm\) and breadth \(1 \rm\,cm\) are removed. To find the area of the remaining card sheet, first find the area of the circular sheet then the area of the two circles of radius \(3.5\rm\, cm\) and a rectangle of length \(3\rm\, cm\) and breadth \(1 \rm\,cm\). Then subtract the area of the three figures from the area of the circular sheet to get the area of the remaining sheet.

Steps:

Radius of circular sheet \((R)\) \(=\) \(14 \rm\,cm\) and radius of smaller circle (r) \(=\) \(3.5\rm\, cm\)

Length of rectangle \((l)\) \(=\) \(3\rm\, cm\) and breadth of rectangle (b) \(=\) \(1 \rm\,cm\)

Area of the remaining sheet = Area of circular sheet - Area of two smaller circle + Area of rectangle

\[\begin{align}&=\! \pi {R^2} \!-\! \left[ {2\pi {r^2} \!+\! \left( {{\rm{Length}} \!\times\! {\rm{Breadth}}} \right)} \right]\\ &=\! \frac{{22}}{7} \!\times\! 14 \!\times\! 14\! - \!\left[ {2 \!\times\! \frac{{22}}{7} \!\times\! 3.5 \!\times\! 3.5 \!+\! \left( {3 \!\times\! 1} \right)} \right]\\ &=\! 22 \!\times\! 2 \!\times\! 14\! -\! \left[ {\frac{{44}}{7} \!\times\! 3.5 \!\times\! 3.5 \!+\! 3} \right]\\ &\!= 616 \!-\! \left( {22 \!\times\! 3.5 \!+\! 3} \right)\\ &=\! 616 \!-\! \left( {77 \!+\! 3} \right)\\ &=\! 616 \!- \!80\\ &= \!536\;\rm c{m^2}\end{align}\]

Therefore, the area of remaining sheet is \(536 \rm\,cm^2.\)