3232.11 – Triangle-ing

In a triangle $ABC$ , $a = 7, b = 8$ and $c = 9.$ Find the altitude of the triangle to side $c$ .

Solution

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

\begin{aligned} 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{aligned}

Therefore,

\begin{aligned} h^2 &=& 7^2 - \frac{11}{3}^2 = \frac{320}{9} \\ h &=& \frac{\sqrt{320}}{3} = 8\frac{\sqrt{5}}{3}. \end{aligned}