Question

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 1

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 \[\dfrac{1}{2} \times \dfrac{{22}}{7} \times {R^2}\].