Question

In the given figure, a circle with cenre O has diameter 112 cm. Two equal circles whose diameter is half the diameter of the bigger circle are drawn inside the circle with centre O. If the shaded portion has been cut-out from the two smaller circles as shown in the figure, then the area of the remaining part of the circle is equal to

• A. 3 times area of shaded figure
• B. 2 times area of shaded figure
• C. area of shaded figure
• D. None of these

Answer 1

The correct option is A

3 times area of shaded figure

Given: the diameter of the circle is $112cm$
$\Rightarrow$ Radius $=\frac{112}{2}=56cm$

$\Rightarrow$ Area of the circle $=\pi \times {56}^2$

Now, if we focus only on one smaller we come to know that the shaded region in both the smaller circle has equal area.

So, we can calculate the area of the shaded region for one smaller circle and multiply it by $2$ to get the area of the whole shaded region.

Area of one shaded region $=\frac{1}{2}\pi (28{)}^2-\frac{1}{2}\pi (14{)}^2+\frac{1}{2}\pi (14{)}^2$

$=\frac{1}{2}\pi (28{)}^2$

$\Rightarrow$ Area of the whole shaded region $=2\times \frac{1}{2}\pi (28{)}^2=\pi (28{)}^2$
Now, area of the remaining circle $=\pi \times (56{)}^2-\pi \times (28{)}^2$

$\frac{\text{Area of the remaining circle}}{\text{Area of the shaded region}}=\frac{\pi \times (56{)}^2-\pi \times (28{)}^2}{\pi \times (28{)}^2}$
$=\frac{\pi \times (56{)}^2}{\pi \times (28{)}^2}-1$
$=\frac{\pi \times 2^2\times (28{)}^2}{\pi \times (28{)}^2}-1$
$=4-1=3$
$\therefore$ Area of the remaining circle $=3\times$ Area of the shaded region