## Trigonometry (11th Edition) Clone

$A = 9.49$
From Exercise 77: $a = \sqrt{34}$ $b = \sqrt{29}$ $c = \sqrt{13}$ We can find the semiperimeter of the triangle: $S = \frac{a+b+c}{2}$ $S = \frac{\sqrt{34}+\sqrt{29}+\sqrt{13}}{2}$ $S = 7.41$ We can use Heron's formula to find the area of the triangle: $A = \sqrt{S(S-a)(S-b)(S-c)}$ $A = \sqrt{7.41(7.41-\sqrt{34})(7.41-\sqrt{29})(7.41-\sqrt{13})}$ $A = \sqrt{90.135}$ $A = 9.49$