In the figure given below, ABCD is a rectangle. AB = 14 cm, BC = 7 cm. From the rectangle, a quarter circle BFEC and a semi-circle DGE are removed. Calculate the area of the remaining piece of the rectangle. (Take \[\pi = \dfrac{{22}}{7}\])
Answer
665.7k+ views
Hint: First of all, find the radius of the quarter circle as its length equals to its smallest side. Then find the radius of the semi-circle as its diameter is equal to the subtraction of radius of the quarter circle from the larger side of the circle. Finally, the area of the remaining piece of the rectangle is given by the area of the rectangle – area of the semi-circle – area of the quarter circle.
Complete step by step solution:
Given that AB = 14 cm and BC = 7 cm
We know that the area of the rectangle of height \[h\] and base \[b\] is given by \[h \times b\].
So, area of rectangle ABCD \[ = 14 \times 7 = 98{\text{ c}}{{\text{m}}^2}\]
From the given figure, clearly radius of the quarter circle BFEC is 7 cm
We know that the area of the quarter circle of radius \[r\] is given by \[\dfrac{1}{4} \times \dfrac{{22}}{7} \times {r^2}\].
So, area of the quarter circle BEFC \[ = \dfrac{1}{4} \times \dfrac{{22}}{7} \times {\left( 7 \right)^2} = 38.5{\text{ c}}{{\text{m}}^2}\]
From the given figure, we have
Diameter of semi-circle DGE + radius of quarter circle BEFC = 14 cm
Diameter of semi-circle BEFC + 7 cm = 14 cm
Therefore, diameter of semi-circle BEFC = 14 cm – 7 cm = 7 cm.
We know that the radius of a semicircle is given by half of its diameter.
So, radius of the semi-circle BEFC \[ = \dfrac{7}{2}{\text{ cm}}\]
We know that the area of a circle of radius \[R\] is given by \[\dfrac{1}{2} \times \dfrac{{22}}{7} \times {R^2}\].
So, area of the semi-circle BEFC \[ = \dfrac{1}{2} \times \dfrac{{22}}{7} \times {\left( {\dfrac{7}{2}} \right)^2} = 19.25{\text{ c}}{{\text{m}}^2}\]
The area of the remaining piece of rectangle = area of the rectangle ABCD – area of the semi-circle
BEFC – area of the quarter circle DGE
\[
= 98 - 19.25 - 38.5 \\
= 40.25{\text{ c}}{{\text{m}}^2} \\
\]
Thus, the area of the remaining piece of the rectangle is \[40.25{\text{ c}}{{\text{m}}^2}\].
Note: The area of the quarter circle of radius \[r\] is given by \[\dfrac{1}{4} \times \dfrac{{22}}{7} \times {r^2}\]. The area of the rectangle of height \[h\] and base \[b\] is given by \[h \times b\]. The area of a circle of radius \[R\] is given by \[\dfrac{1}{2} \times \dfrac{{22}}{7} \times {R^2}\].
Complete step by step solution:
Given that AB = 14 cm and BC = 7 cm
We know that the area of the rectangle of height \[h\] and base \[b\] is given by \[h \times b\].
So, area of rectangle ABCD \[ = 14 \times 7 = 98{\text{ c}}{{\text{m}}^2}\]
From the given figure, clearly radius of the quarter circle BFEC is 7 cm
We know that the area of the quarter circle of radius \[r\] is given by \[\dfrac{1}{4} \times \dfrac{{22}}{7} \times {r^2}\].
So, area of the quarter circle BEFC \[ = \dfrac{1}{4} \times \dfrac{{22}}{7} \times {\left( 7 \right)^2} = 38.5{\text{ c}}{{\text{m}}^2}\]
From the given figure, we have
Diameter of semi-circle DGE + radius of quarter circle BEFC = 14 cm
Diameter of semi-circle BEFC + 7 cm = 14 cm
Therefore, diameter of semi-circle BEFC = 14 cm – 7 cm = 7 cm.
We know that the radius of a semicircle is given by half of its diameter.
So, radius of the semi-circle BEFC \[ = \dfrac{7}{2}{\text{ cm}}\]
We know that the area of a circle of radius \[R\] is given by \[\dfrac{1}{2} \times \dfrac{{22}}{7} \times {R^2}\].
So, area of the semi-circle BEFC \[ = \dfrac{1}{2} \times \dfrac{{22}}{7} \times {\left( {\dfrac{7}{2}} \right)^2} = 19.25{\text{ c}}{{\text{m}}^2}\]
The area of the remaining piece of rectangle = area of the rectangle ABCD – area of the semi-circle
BEFC – area of the quarter circle DGE
\[
= 98 - 19.25 - 38.5 \\
= 40.25{\text{ c}}{{\text{m}}^2} \\
\]
Thus, the area of the remaining piece of the rectangle is \[40.25{\text{ c}}{{\text{m}}^2}\].
Note: The area of the quarter circle of radius \[r\] is given by \[\dfrac{1}{4} \times \dfrac{{22}}{7} \times {r^2}\]. The area of the rectangle of height \[h\] and base \[b\] is given by \[h \times b\]. The area of a circle of radius \[R\] is given by \[\dfrac{1}{2} \times \dfrac{{22}}{7} \times {R^2}\].
Recently Updated Pages
A Paragraph on Pollution in about 100-150 Words

What is BLO What is the full form of BLO class 8 social science CBSE

10 examples of friction in our daily life

Draw a diagram of nephron and explain its structur class 11 biology CBSE

Write structures of the following compounds i 2 Chloro3methylpentane class 11 chemistry CBSE

XIX+XXX A 49 B 51 C 55 D 44 class 5 maths CBSE

Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE

1 GB equals how many MB?

10 examples of evaporation in daily life with explanations

What is the full form of POSCO class 10 social science CBSE

Which is the hottest planet in the Solar system A Earth class 10 social science CBSE

Name any four life processes in living things class 10 biology CBSE

