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In triangle FGH, begin mathsize 14px style GJ with bar on top end style is an angle bisector of begin mathsize 14px style angle G end style and perpendicular to begin mathsize 14px style FH with bar on top end style. What is the length of begin mathsize 14px style FH with bar on top end style?

User Marged
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The length of FH is 16 units.

In triangle FGH, GJ is an angle bisector of angle G and perpendicular to FH. We are given that FG = 3x - 8, FJ = x, and GH = 16. We need to find the length of FH.

Since GJ is an angle bisector of angle G, it divides angle G into two congruent angles, ∠FJG and ∠HJG. Therefore, triangles FJG and HJG are right triangles and share a hypotenuse, GJ.

We can use the Pythagorean theorem to find the lengths of FJ and FH. In triangle FJG, we have:

FJ^2 + JG^2 = FG^2

Substituting the given values, we get:

x^2 + JG^2 = (3x - 8)^2

Solving for JG, we get:

JG = √(14x^2 - 48x + 64)

Similarly, in triangle HJG, we have:

JH^2 + JG^2 = GH^2

Substituting the given values, we get:

JH^2 + √(14x^2 - 48x + 64)^2 = 16^2

Solving for JH, we get:

JH = √(256 - 14x^2 + 48x - 64)

Since FH is the hypotenuse of right triangles FJG and HJG, we can find its length using the Pythagorean theorem:

FH^2 = FJ^2 + JH^2

Substituting the expressions we obtained for FJ and JH, we get:

FH^2 = x^2 + √(14x^2 - 48x + 64)^2 + √(256 - 14x^2 + 48x - 64)^2

Simplifying the expression, we get:

FH^2 = 256

Therefore, the length of FH is 16 units.

User Ben Francom
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