2006 iTest Problems/Problem 22

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Problem

Triangle $ABC$ has sidelengths $AB=75$, $BC=100$, and $CA=125$. Point $D$ is the foot of the altitude from $B$, and $E$ lies on segment $BC$ such that $DE\perp BC$. Find the area of the triangle $BDE$.

[asy] import olympiad; size(170); defaultpen(linewidth(0.7)+fontsize(11pt)); pair A = origin, B = (9,12), C = (25,0), D = foot(B,A,C), E = foot(D,B,C); draw(A--B--C--A^^B--D--E); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$D$",D,S); label("$E$",E,NE); [/asy]

Solution

Note that because the area of a triangle is the same, the product of the right triangle's legs equals the hypotenuse multiplied by the altitude to the hypotenuse. Thus, we have $75 \cdot 100 = 125 \cdot BD$, so $BD = 60$. We also have $60 \cdot 80 = 100 \cdot DE$, so $DE = 48$.


By the Pythagorean Theorem, $BE = \sqrt{60^2 - 48^2} = 36$. Therefore, the area of triangle $BDE$ equals $\tfrac12 \cdot 36 \cdot 48 = \boxed{864}$.

See Also

2006 iTest (Problems)
Preceded by:
Problem 21
Followed by:
Problem 23
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