Question

A paper is in the form of a rectangle ABCD, in which AB = 16 cm and BC = 12 cm. A semicircular portion with BC as diameter is cut-off. Find the area of remaining part.

Answer 1

Given that, a paper is in the form a rectangle $ABCD$, where $AB=16cm$ and $BC=12cm$.

Also, a semicircle with diameter $12cm$ is cut from the rectangle.

To find out: The area of the remaining part.

Area of the remaining part $=$ Area of the rectangle $-$ Area of the semicircle

We know that, area of a rectangle $=l\times b$

Hence, area of the given rectangle $=16\times 12=192c{m}^2$

We also know that, area of a semicircle $=\frac{1}{2}\times$area of a circle $=\frac{1}{2}\pi r^2$

The radius of the semicircle is $6cm$ $[r=\frac{12}{2}]$

Hence, area of the semicircle $=\frac{1}{2}\times \frac{22}{7}\times 6\times 6$ $[\text{Using}\pi =\frac{22}{7}]$

$\therefore$ Area of the semicircle $=\frac{396}{7}c{m}^2$

So, area of the remaining part $=192-\frac{396}{7}=\frac{1344-396}{7}$

$\Rightarrow \frac{948}{7}=135.43c{m}^2$

Hence, the area of the remaining part is $135.43c{m}^2$.