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

*h*^{2} = 7^{2} – *x*^{2} = 8^{2} – (9 – *x*)^{2}

7^{2} – *x*^{2} = 8^{2} – (9^{2} – 18*x* + *x*^{2}) → 7^{2} – *x*^{2} = 8^{2} – 9^{2} + 18*x* – *x*^{2}

49 = 64 – 81 + 18*x* → 18*x* = 66 → *x* = 11/3.

*h*^{2} = 7^{2} – (11/3)^{2} = 49 – 121/9 = (441 – 121)/9 = 320/9 = (64/9) ⋅ 5

*h* = 8√5 / 3