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.
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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