Question

6. a semicircle has as its diameter the hypotenuse of a right triangle shown below. determine the area of the semicircle to the nearest tenth of a square centimeter. show how you arrived at your answer.

Answer 1

Answer:

$A = 137.3cm^2$

Step-by-step explanation:

Given

See attachment

Required

The area of the semicircle

First, we calculate the hypotenuse (h) of the triangle

Considering only the triangle, we have:

$\cos(68) = \frac{7}{h}$ --- cosine formula

Make h the subject

$h = \frac{7}{\cos(68)}$

$h = \frac{7}{0.3746}$

$h = 18.7$

The area of the semicircle is then calculated as:

$A = \frac{\pi h^2}{8}$

This gives:

$A = \frac{3.14 * 18.7^2}{8}$

$A = \frac{1098.03}{8}$

$A = 137.3cm^2$