Rectangle of the legs and rectangle of hypotenuse and altitude

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Let $KLM$ be a right triangle with the right angle at $L$, and let $h$ be the altitude to the hypotenuse $KM$.

• (a) Prove that the rectangle whose sides are the two legs $LK$ and $LM$ is equivalent to the rectangle whose sides are the hypotenuse $KM$ and the altitude $h$.
• (b) Given $\overline{LK}=9$ and $\overline{LM}=12$, compute the hypotenuse $\overline{KM}$ and the altitude $h$.

Solution

(a) Proof. The area of triangle $KLM$ can be written in two ways. Taking the two legs (perpendicular to each other) as base and height: $\mathcal{A}=\frac{\overline{LK}\cdot\overline{LM}}{2}.$ Taking instead the hypotenuse $KM$ as base and the segment $h$ relative to it as height: $\mathcal{A}=\frac{\overline{KM}\cdot h}{2}.$ Equating the two expressions: $\overline{LK}\cdot\overline{LM}=\overline{KM}\cdot h.$ The rectangle of the legs and the rectangle of hypotenuse and altitude thus have the same area: they are equivalent. $\;\blacksquare$

(b) Numerical computation. By the Pythagorean theorem: $\overline{KM}=\sqrt{\overline{LK}^2+\overline{LM}^2}=\sqrt{9^2+12^2}=\sqrt{81+144}=\sqrt{225}=15.$ From the relation proved in part (a): $h=\frac{\overline{LK}\cdot\overline{LM}}{\overline{KM}}=\frac{9\cdot 12}{15}=\frac{108}{15}=7{,}2.$ $\boxed{\overline{KM}=15,\qquad h=7{,}2}$