Question

In the given figure, ABCD is a rectangle with AB = 14 cm and BC = 7 cm. Taking DC, BC and AD as diameters, three semicircles are drawn as shown in the figure. Find the area of the shaded region.

Answer 1

Hint: In order to find the area of the shaded region find the area of each of the geometrical figures separately with the help of their formulas. Then find the area of the shaded region by adding some area and subtracting some of them.

Complete step-by-step answer:
Given that $AB = 14cm,BC = AD = 7cm$
As we know that area of the rectangle $ = l \times b$
Area of the semicircle $ = \dfrac{{\pi r{}^2}}{2}$
Here the radius of bigger semicircle $\left( {{r_1}} \right) = \dfrac{{14}}{2} = 7cm$
And the radius of smaller semicircle $\left( {{r_2}} \right) = \dfrac{7}{2}cm$
As it is clear from the figure that

Area of the shaded region = Area of rectangle ABCD + 2 x (Area of small semicircle) – area of big semicircle.
So area of shaded region:
\[ = \left( {l \times b} \right) + 2 \times \left( {\dfrac{{\pi r_2^2}}{2}} \right) - \left( {\dfrac{{\pi r_1^2}}{2}} \right)\]
Now let us substitute the values in the above formula.
Area of shaded region:
\[
   = \left( {14 \times 7} \right) + 2 \times \left( {\dfrac{{\pi {{\left( {\dfrac{7}{2}}

\right)}^2}}}{2}} \right) - \left( {\dfrac{{\pi {{\left( 7 \right)}^2}}}{2}} \right) \\


   = \left( {14 \times 7} \right) + \left( {2 \times \dfrac{1}{2} \times \dfrac{{22}}{7} \times

\dfrac{7}{2} \times \dfrac{7}{2}} \right) - \left( {\dfrac{1}{2} \times \dfrac{{22}}{7} \times 7

\times 7} \right) \\

   = 98 + \dfrac{{77}}{2} - 77 \\

   = 98 + 38.5 - 77 \\

   = 59.5c{m^2} \\

 \]

Hence, the area of the shaded region is \[59.5c{m^2}\] .

Note: In order to solve such types of problems, students must first recognize the area of the shaded region as a sum and difference of some geometrical figures. As it is easier to find the areas of the geometrical figure than random figures by the help of formulas. Students must remember the formulas of the area of some common figures.