#### Answer

$A = 9.49$

#### Work Step by Step

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:
$A = \sqrt{S(S-a)(S-b)(S-c)}$
$A = \sqrt{7.41(7.41-\sqrt{34})(7.41-\sqrt{29})(7.41-\sqrt{13})}$
$A = \sqrt{90.135}$
$A = 9.49$