# 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 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • U1 • U2 • U3 • U4 • U5 • U6 • U7 • U8 • U9 • U10
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