Let $F$, $D$, and $E$ be points on the sides $[AB]$, $[BC]$, and $[CA]$ of $\triangle ABC$, respectively, such that $\triangle DEF$ is an isosceles right triangle with hypotenuse $[EF]$. The altitude of $\triangle ABC$ passing through $A$ is $10$ cm. If $|BC|=30$ cm, and $EF \parallel BC$, calculate the perimeter of $\triangle DEF$.
Problem
Source:
Tags: geometry, perimeter
admin25
19.01.2013 23:55
Let $ X $ be the foot of the altitude from $ A $ to $ BC $, and let $ Y $ be the foot of the altitude from $ A $ to $ EF $. Since $ EF||BC $, $ A, Y, X $ are collinear. Let $ AY=x $, so $ YX=10-x $ is the height of $ \triangle DEF $ from $ D $. Since $ \triangle DEF $ is an isosceles right triangle, its hypotenuse is twice the height to its hypotenuse, so $ FE=20-2x $.
now, since $ EF||BC $, $ \triangle AFE\sim\triangle ABC $, so $ \frac{x}{10}=\frac{20-2x}{30} $, and so $ x=4 $. Therefore, the height of the isosceles right triangle is $ 10-4=6 $, and so it has side lengths $ 6\sqrt{2}, 6\sqrt{2}, 12 $, and its perimeter is $ \boxed{12\sqrt{2}+12} $.
OlympusHero
21.10.2021 23:06
Denote $FD=ED=x$, so $EF=x\sqrt2$. Then the ratio of similarity of $\triangle AFE$ and $\triangle ABC$ is $\frac{x\sqrt2}{30}$, so the altitude from $A$ to $FE$ has length $\frac{x\sqrt2}{3}$. The altitude from $D$ to $FE$ has length $\frac{x\sqrt2}{2}$, so $\frac{x\sqrt2}{3}+\frac{x\sqrt2}{2}=10 \implies x = 6\sqrt2$. We want $x+x+x\sqrt2 = \boxed{12\sqrt2+12}$.