Question

# A paper is in the form of a rectangle ABCD where AB = 22cm and BC = 14cm. A semi-circle portion with BC as diameter is cut off. Find the area of the remaining part. $\left( \pi =\dfrac{22}{7} \right)$

Hint: Area of a rectangle can be given by relation length $\times$ breadth and area of circle is given as $\pi {{r}^{2}}$ where r is radius and value of $\pi$ is $\dfrac{22}{7}$. Now, find the remaining area by calculating the difference of the areas of rectangle and semi-circle.

Here, we have a rectangle (paper) ABCD where sides AB = 22cm and BC = 14cm, and we have to determine the area of portion remaining after cutting a semi-circular portion with BC as a diameter. Hence, diagram can be given as

Here O is the center of the semi-circle and mid-point of BC as well.
Hence, radius of semicircle BC is 7cm, i.e., $\left( \dfrac{14}{2} \right)$.
Now, we can get the area of the remaining portion of the rectangle by subtracting the area of the whole rectangle by the area of the semi-circle.
Now, we know
\begin{align} & \text{area of rectangle = length }\!\!\times\!\!\text{ breadth} \\ & \text{area of circle = }\pi {{\text{r}}^{2}} \\ \end{align}
Where r = radius, $\pi =\dfrac{22}{7}$
Hence, area of semicircle $=\dfrac{\pi {{r}^{2}}}{2}$
So, we can get area of remaining part from the given diagram as
Remaining area = $AB\times BC-\dfrac{\pi {{r}^{2}}}{2}$
$=22\times 14-\dfrac{22}{7}\times \dfrac{1}{2}\times {{\left( 7 \right)}^{2}}$
\begin{align} & =308-11\times 7 \\ & =308-77 \\ & =231c{{m}^{2}} \\ \end{align}
Hence, the area of the remaining part can be given as 231 sq. cm.
So, option (B) is the correct answer.

Note:
One may get confused between the formula of area and circumference of a circle. So, one may go wrong while using the area of the circle as $2\pi r$. Hence, be clear with the formula of areas of standard shapes.
One may get confused with the center of the semicircle that is the midpoint of BC. Itâ€™s because BC is the diameter of a semi-circle and the midpoint of diameter is center.