Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 7 - Applications of Trigonometry and Vectors - Section 7.3 The Law of Cosines - 7.3 Exercises - Page 324: 68


$228$ yd$^{2}$

Work Step by Step

Since the lengths of three sides of the triangle are given, we can use the Heron's formula to find the area of the triangle. To use the Heron's formula, we first need to find the semi-perimeter $s$. We substitute the values of the sides of the triangle in the formula below to find the semi-perimeter $s$: $s=\frac{1}{2}(a+b+c)$ $s=\frac{1}{2}(25.4+38.2+19.8)$ $s=\frac{1}{2}(83.4)$ $s=41.7$ Now, we use the Heron's formula to find the area of the triangle: $A=\sqrt{s(s-a)(s-b)(s-c)}$ $A=\sqrt{41.7(41.7-25.4)(41.7-38.2)(41.7-19.8)}$ $A=\sqrt{41.7(16.3)(3.5)(21.9)}$ $A=\sqrt{52099.77}$ $A=228.25\approx228$ Therefore, the area of the triangle is $228$ yd$^{2}$.
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