Question

What is the area of △FGH to the nearest tenth of a square meter? The image is of a triangle GHF with base GH length 2m, FG is 2.5 m and angle FGH is 121 degrees.

Answer 1

First, we are going to use the law of cosines to find the length of the line segment FH:
$FH= \sqrt{2.5^{2}+2^{2}-(2)(2.5)Cos(121)}$
$FH= \sqrt{2.5^{2}+2^{2}-5Cos(121)}$
$FH=3.2$

Next, we are going to use the semi-perimeter formula: $s= \frac{GH+FG+FH}{2}$
$s= \frac{2+2.5+3.2}{2}$
$s= \frac{7.7}{2}$
$s=3.9$

Now that we have the semi-perimeter of our triangle, we can find its area using Heron's formula:
$A= \sqrt{s(s-GH)(s-FG)(s-FH)}$
$A= \sqrt{3.9(3.9-2)(3.9-2.5)(3.9-3.2)}$
$A= \sqrt{3.9(1.9)(1.4)(0.7)}$
$A=2.7m^{2}$

We can conclude that the area of the triangle GHF is 2.7 $m^{2}$.