Track / Stage 6 / 109 of 400 #1109 of 1964

Problem 1109

National olympiad, first round

Geometry Difficulty 6.1 Prove it

Prove that the area of a parallelogram is equal to the product of its two heights divided by the sine of the angle between them, i.e.,

$S=\frac{h_{a} h_{b}}{\sin \gamma}$

where $h_{\mathrm{a}}$ and $h_{\mathrm{b}}$ are the heights dropped to the adjacent sides, equal to $a$ and $b$ , and $\gamma$ is the angle between these sides.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $B M=h_{\mathrm{b}}$ and $D N=h_{\mathrm{a}}$ be the heights of the parallelogram $A B C D$ , dropped to the sides $A D$ and $A B$ respectively, $A B=a$ and $A D=b-$ be the sides of the parallelogram, $\gamma$ be the angle between the lines $B M$ and $D N$ . Then the angle between the lines $A B$ and $A D$ is also $\gamma$ . From the right triangles $A N D$ and $B M A$ , we find that

$b=A D=\frac{D N}{\sin \angle B A D}=\frac{h_{a}}{\sin \gamma}, a=A B=\frac{B M}{\sin \angle B A D}=\frac{h_{b}}{\sin \gamma}$

Therefore,

$S_{\mathrm{ABCD}}=A B \cdot A D \cdot \sin \angle B A D=a b \sin \gamma=\frac{h_{b}}{\sin \gamma} \cdot \frac{h_{a}}{\sin \gamma} \cdot \sin \gamma=\frac{h_{a} h_{b}}{\sin \gamma}$

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.