Hypotenuse, altitude and projections of the legs

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Triangle $RST$ is right-angled at $S$ and has legs $\overline{SR}=15$ and $\overline{ST}=20$. Let $H$ be the foot of the altitude to the hypotenuse $RT$. Compute:

• the hypotenuse $\overline{RT}$;
• the altitude $\overline{SH}$ to the hypotenuse;
• the two projections $\overline{RH}$ and $\overline{TH}$ of the legs onto the hypotenuse.

Solution

Hypotenuse. By the Pythagorean theorem: $\overline{RT}=\sqrt{\overline{SR}^2+\overline{ST}^2}=\sqrt{15^2+20^2}=\sqrt{225+400}=\sqrt{625}=25.$ Altitude to the hypotenuse. Equating the two expressions for the area ($\tfrac12\,\overline{SR}\cdot\overline{ST}=\tfrac12\,\overline{RT}\cdot\overline{SH}$): $\overline{SH}=\frac{\overline{SR}\cdot\overline{ST}}{\overline{RT}}=\frac{15\cdot 20}{25}=\frac{300}{25}=12.$ Projections of the legs. By Euclid’s first theorem each leg is the geometric mean between the hypotenuse and its own projection, so $\overline{RH}=\dfrac{\overline{SR}^2}{\overline{RT}}$ and $\overline{TH}=\dfrac{\overline{ST}^2}{\overline{RT}}$: $\overline{RH}=\frac{15^2}{25}=\frac{225}{25}=9,\qquad \overline{TH}=\frac{20^2}{25}=\frac{400}{25}=16.$ Check: $\overline{RH}+\overline{TH}=9+16=25=\overline{RT}$ and $\overline{SH}^2=12^2=144=9\cdot 16=\overline{RH}\cdot\overline{TH}$ (Euclid’s second theorem). $\boxed{\overline{RT}=25,\quad \overline{SH}=12,\quad \overline{RH}=9,\quad \overline{TH}=16}$