Need help ASAPWhat is the area of this triangleEnter your answer as a decimal in the box

Accepted Solution

Answer:[tex]\boxed{A=43.54cm^2}[/tex]Step-by-step explanation:To find this area we will use the law of cosine and the Heron's formula. First of all, let't find the unknown side using the law of cosine:[tex]x^2=12^2+8^2-2(12)(8)cos(65^{\circ}) \\ \\ x^2=144+64-192(0.42) \\ \\ x^2=208-80.64 \\ \\ x^2=127.36 \\ \\ x=\sqrt{127.36} \\ \\ \therefore \boxed{x=11.28cm}[/tex]Heron's formula (also called hero's formula) is used to find the area of a triangle using the triangle's side lengths and the semiperimeter. A polygon's semiperimeter s is half its perimeter. So the area of a triangle can be found by:[tex]A=\sqrt{s(s-a)(s-b)(s-c)}[/tex] being [tex]a,\:b\:and\:c[/tex] the corresponding sides of the triangle.So the semiperimeter is:[tex]s=\frac{12+8+11.28}{2} \\ \\ s=15.64cm[/tex]So the area is:[tex]A=\sqrt{15.64(15.64-12)(15.64-8)(15.64-11.28)} \\ \\ \therefore \boxed{A=43.54cm^2}[/tex]