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: 37

Answer

$A=96.0~cm^{2}$ $P=48.0~cm$

Work Step by Step

To find the area, we use Heron's formula. We have the sides: $a=12cm$ $b=16cm$ $c=20cm$ Thus: $s=\frac{1}{2}$$(a+b+c)=\frac{1}{2}(12+16+20)$ $s=24cm$ and the area is: $A=\sqrt {s(s-a)(s-b)(s-c)}$ Thus: $A=\sqrt {24cm(24cm-12cm)(24cm-16cm)(24cm-20cm)}$ $A=96.0~cm^{2}$ To find the perimeter, we add the lengths of all the sides: $P=12cm+16cm+20cm=48.0~cm$
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