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. $\left( \pi =\dfrac{22}{7} \right)$
Last updated date: 20th Mar 2023
•
Total views: 303.9k
•
Views today: 4.83k
Answer
303.9k+ views
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.
Complete step-by-step answer:
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. It’s because BC is the diameter of a semi-circle and the midpoint of diameter is center.
Complete step-by-step answer:
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. It’s because BC is the diameter of a semi-circle and the midpoint of diameter is center.
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE
