Question

The figure below shows two half-circles at the ends of a rectangle with the dimensions shown.

Answer 1

Answer:

$A_T=94.63in^2$

Explanation We need to find the are of the figure provided, the area is composed of three parts, rectangle, and two half circles: therefore the total area would be the sum of the three:

Area Rectangle:

$A_R=w\times l=15in\times5in=75in^2$

Area of two half circles:

$\begin{gathered} A_c=\pi r^2_{} \\ r=\frac{5}{2}in=2.5in \\ \therefore\rightarrow \\ A_c=\pi(2.5in)^2=(3.141\times6.25)in^2=19.63in^2 \\ \therefore\rightarrow \\ A_{C.H}=\frac{19.63}{2}in^2=9.82in^2 \\ \text{ SInce We have two such halves, therefore the total are of the two is} \\ A_{C-T}=2\times9.82in^2=19.63in^2 \end{gathered}$

The total area of the figure is:

$\begin{gathered} A_T=A_R+A_{C-T}=75in^2+19.63in^2=94.63in^2 \\ A_T=94.63in^2 \end{gathered}$