Question

Triangle FGH, with vertices F(2,-8), G(7,-7), and H(6,-4), is drawn inside a rectangle, as shown below.

Answer 1

The area of the triangle FGH is 13√10 unit²

Given,

The vertices;

F(2, -8), G(7, -7), H(6, -4)

We have to find the area of the triangle.

Area of triangle = 1/2bh²

Here,

Area of triangle FGH = 1/2 (GH) (FG)²

Now,

Length of FG = $\sqrt{(7-2)^{2} =(-8--7)^{2} }$  = $\sqrt{5^{2}+1^{2} }$  = $\sqrt{25+1}$ = √26

Length of GH = $\sqrt{(-7--4)^{2}+(7-6)^{2} }$ = $\sqrt{-3^{2}-1^{2} }$ = $\sqrt{9+1}$ = √10

Then,

Area of triangle FGH = 1/2 (GH) (FG)²

Area of triangle FGH = 1/2 × √10 × √26²

Area of triangle FGH = 1/2 × √10 × 26

Area of triangle FGH = 13√10

Therefore,

The area of the triangle FGH is 13√10 unit²

Learn more about triangle here;

brainly.com/question/12949215

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