**Question:**

A sheet of paper is in the form of a rectangle ABCD in which AB = 40 cm and AD = 28 cm. A semi-circular portion with BC as diameter is cut off. Find the area of the remaining paper.

**Solution:**

The length and width of rectangle *ABCD* is given by andrespectively.

Now, we will find the area of rectangle.

Area of rectangle $=l \times w$

$=40 \times 28$

$=1120 \mathrm{~cm}^{2}$

It is given that a semicircular portion with BC as diameter is cutoff from rectangle. So,

radius of semicircle $=\frac{B C}{2}$

$=\frac{28}{2}$

$=14 \mathrm{~cm}$

Now, The area of semicircle $=\frac{1}{2} \pi r^{2}$

Substituting the value of *r*,

The area of semicircle $=\frac{1}{2} \times \frac{22}{7} \times 14 \times 14$

$=308 \mathrm{~cm}^{2}$

The area A of remaining paper is

$A=$ Area of rectangle $-$ Area of semicircle

$=1120-308$

$=812 \mathrm{~cm}^{2}$

Thus, the area of remaining paper is $812 \mathrm{~cm}^{2}$.

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