Question

9 in. 12 in. perimeter (terms of pi) __________ perimeter (number) __________

Explanation:

Step1: Find the length of the hypotenuse of the right - triangle

Use the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(a = 9\) and \(b = 12\).
\(c=\sqrt{9^{2}+12^{2}}=\sqrt{81 + 144}=\sqrt{225}=15\) in.

Step2: Find the length of the arc of the semi - circle

The diameter of the semi - circle is the hypotenuse \(c = 15\) in, so the radius \(r=\frac{15}{2}\) in. The formula for the length of an arc of a semi - circle is \(l=\frac{1}{2}\times2\pi r=\pi r\). Substituting \(r = \frac{15}{2}\), we get \(l=\frac{15\pi}{2}\) in.

Step3: Calculate the perimeter in terms of \(\pi\)

The perimeter \(P\) of the figure is the sum of the two legs of the right - triangle and the arc of the semi - circle. \(P=9 + 12+\frac{15\pi}{2}=21+\frac{15\pi}{2}\) in.

Step4: Calculate the perimeter as a number

Substitute \(\pi\approx3.14\) into the perimeter formula. \(P=21+\frac{15\times3.14}{2}=21 + 23.55=44.55\) in.

Answer:

Perimeter (terms of Pi): \(21+\frac{15\pi}{2}\) in
Perimeter (number): \(44.55\) in