Question

a rectangular paperboard measuring 32 in long and 19 in wide has a semicircle cut out of it, as shown below. what is the perimeter of the paperboard that remains after the semicircle is removed? (use the value 3.14 for π, and do not round your answer. be sure to include the correct unit in your answer.)

Explanation:

Step1: Calculate the perimeter of the rectangle part

The formula for the perimeter of a rectangle is \(P = 2(l + w)\). But here, one side (the side where the semicircle is removed) is not included. The two lengths are \(32\) in each, and one width is \(19\) in. So the sum of these parts is \(32+32 + 19=83\) in.

Step2: Calculate the length of the semicircle arc

The formula for the circumference of a full - circle is \(C = 2\pi r\). For a semicircle, the length of the arc is \(C_{semicircle}=\pi r\). Given the diameter of the semicircle is \(19\) in (so the radius \(r=\frac{19}{2}\) in), and using \(\pi = 3.14\), the length of the semicircle arc is \(3.14\times\frac{19}{2}=3.14\times9.5 = 29.83\) in.

Step3: Calculate the total perimeter

The total perimeter \(P\) of the paperboard is the sum of the rectangle parts and the semicircle arc. So \(P=83 + 29.83=112.83\) in.

Answer:

\(112.83\) in