## Trigonometry (11th Edition) Clone

$A = 24\sqrt{3}$ We also find this area when we use Heron's formula.
We can find the area of the triangle: $A = \frac{1}{2}bh$ $A = \frac{1}{2}(16)(3\sqrt{3})$ $A = 24\sqrt{3}$ We can find the semiperimeter of the triangle: $S = \frac{a+b+c}{2}$ $S = \frac{6+14+16}{2}$ $S = 18$ We can use Heron's formula to find the area of the triangle: $A = \sqrt{S(S-a)(S-b)(S-c)}$ $A = \sqrt{18(18-6)(18-14)(18-16)}$ $A = \sqrt{1728}$ $A = 24\sqrt{3}$