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: 29

Answer

$158\ m^2, \ 60.8\ m$

Work Step by Step

Since the right triangle is isosceles, then the legs are equal. The area is given by $$ Area= \frac{1}{2} base \times height=\frac{1}{2}\times17.8\times 17.8=158.42\ m^2\approx 158~m^2 .$$ The perimeter of a triangle is the sum of the lengths of the sides: $$perimeter=17.8+17.8+17.8\sqrt{2}=60.8\ m .$$ Where we used the Pythagorean theorem to find the hypotenuse: $c^2=a^2+b^2$ $c=\sqrt{17.8^2+17.8^2}=17.8\sqrt{2}$
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