Chapter 10 - Geometry - 10.4 Area and Circumference - Exercise Set 10.4 - Page 648: 36

The perimeter of the figure is \[285.6\text{ m}\] and the area is\[4456\text{ }{{\text{m}}^{2}}\].

The perimeter of any figure is the sum of all the sides that makes up the boundary of the figure. The given figure consists of two semicircles attached on the breadth of the rectangle. The perimeter will be equal to the sum of twice the length of the rectangle and twice the circumference of a semicircle. Twice the circumference of the semicircle will be equal to circumference of a circle. The circumference of the circle with diameter equal to the breadth of the rectangle\[\left( 40\text{ m} \right)\]is as follows: \[\begin{align} & C=\pi d \\ & =40\pi \text{ m} \end{align}\] \[\begin{align} & \text{Perimeter}=C+80+80 \\ & =\left( 40\pi +160 \right)m \\ & =285.6\text{m} \end{align}\] The area of the figure will be the sum of the areas of the rectangle and the two semicircles. The areas of two semicircles will be equal to the area of one circle with diameter equal to \[40\text{m}\](radius will be\[20\text{m}\]). Compute the area of the circle as follows: \[\begin{align} & {{A}_{1}}=\pi {{r}^{2}} \\ & =\text{400}\pi {{\text{m}}^{2}} \end{align}\] The area of the rectangle will be as follows: \[\begin{align} & {{A}_{2}}=lb \\ & =80\text{ m}\times 40\text{ m} \\ & =3200\text{ }{{\text{m}}^{2}} \end{align}\] The total area will be the sum of these two areas: \[\begin{align} & A={{A}_{1}}+{{A}_{2}} \\ & =\left( 400\pi +3200 \right){{m}^{2}} \\ & =4456\text{ }{{\text{m}}^{2}} \end{align}\] Hence, the perimeter of the figure is \[285.6\text{m}\] and the area is\[4456{{\text{m}}^{2}}\].