Question

A rectangular paperboard measuring 29 in long and 21 in wide has a semicircle cut out of it, as shown below. What is the perimeter of the paperboard thatremains after the semicircle is removed? (Use the value 3.14 for Tr, and do not round your answer. Be sure to include the correct unit in your answer.)21 in29 in

Answer 1

Answer:

111.97 in

Explanation:

From the given figure, we can deduce the following

Length of the rectangular paperboard = 29 in

Width of the rectangular paperboard = 21 in

The radius(r) of the semicircle (half of the width of the paperboard) = 21/2 = 10.5 in

Let's go ahead and determine the circumference(C) of the semicircle as shown below;

$\begin{gathered} C=\pi r \\ =3.14\cdot10.5 \\ =32.97\text{ in} \end{gathered}$

We can now go ahead and determine the perimeter of the paperboard that remains after the semicircle is removed;

$\text{Perimeter}=21+29+29+32.97=111.97\text{ in}$

Therefore, the perimeter of the paperboard that remains after the semicircle is removed is 111.97 in.