3232.11 – Triangle-ing

In a triangle △ABC \triangle ABC with side lengths a=7,b=8a = 7, b = 8 and c=9c = 9, find the altitude of the triangle to side cc.

Solution

Please refer to the diagram below. Ready? Here we go.

h2=72−x2=82−(9−x)249−x2=64−81+18x−x249=18x−1718x=66x=11/3.\begin{align*}h^2 &&= 7^2 - x^2 &= 8^2 - (9 - x)^2 \\&& 49 - x^2 &= 64 - 81 + 18 x - x^2 \\&& 49 &= 18x - 17 \\&& 18x &= 66 \\&& x &= 11/3.\end{align*}

Therefore,

h2=72−1132=3209h=3203=853.\begin{align*}h^2 &= 7^2 - \dfrac{11}{3}^2 = \dfrac{320}{9} \\h &= \dfrac{\sqrt{320}}{3} = 8\dfrac{\sqrt{5}}{3}.\end{align*}