## Trigonometry (11th Edition) Clone

Published by Pearson

# Chapter 7 - Applications of Trigonometry and Vectors - Section 7.3 The Law of Cosines - 7.3 Exercises - Page 324: 64

#### Answer

$A = 15\sqrt{3}$ We also find this area when we use Heron's formula.

#### Work Step by Step

We can find the area of the triangle: $A = \frac{1}{2}bh$ $A = \frac{1}{2}(10)(3\sqrt{3})$ $A = 15\sqrt{3}$ We can find the semiperimeter of the triangle: $S = \frac{a+b+c}{2}$ $S = \frac{6+10+14}{2}$ $S = 15$ We can use Heron's formula to find the area of the triangle: $A = \sqrt{S(S-a)(S-b)(S-c)}$ $A = \sqrt{15(15-6)(15-10)(15-14)}$ $A = \sqrt{675}$ $A = 15\sqrt{3}$

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