Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 12 - Section 12.3 - Triangles - Exercise - Page 400: 30

Answer

$147~ft^2\ , \ 56.0\ ft$

Work Step by Step

If we divide the triangle into two parts, we have two right triangles. Recall the Pythagorean theorem: $$c^2=a^2+b^2.$$ We can use this to calculate the height of the triangle by making one of the legs equal to $16\div 2=8$: $h^2+8^2=20^2$ $h=\sqrt{20^2-8^2}=18.33$. Now that we have the height, we calculate the area: $$ Area= \frac{1}{2} base \times height=\frac{1}{2}\times16\times 18.33=146.64\ ft^2\approx 147~ft^2 .$$ The perimeter of the triangle is the sum of the lengths of the sides. Thus, we have $$perimeter= 16+20+20=56.0\ ft .$$
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