Question

What is the volume in cubic inches of the solid figure, rounded to the nearest cubic inch? Do not use units or commas in your answer.

Answer 1

Answer:

1131 cubic inches.

Step-by-step explanation:

The front side of the figure contains a rectangle and a semicircle.

Area of rectangle is

$A_1=length\times breadth$

$A_1=11\times 12$

$A_1=132\text{ in}^2$

Radius of semicircle is

$r=17-11=6\text{ in}$

Area of semicircle is

$A_2=\dfrac{1}{2}\pi r^2$

$A_2=\dfrac{1}{2}\pi (6)^2$

$A_2\approx 56.55$

Area of front side is

$A=A_1+A_2=132+56.55=188.55\text{ in}^2$

Let front side is the base of prism and height is 6 in. So, volume of given figure is

$V=\text{Base area}\times height$

$V=188.55\times 6$

$V=1131.3$

$V\approx 1131\text{ in}^3$

Therefore, the required volume is 1131 cubic inches.