# 1995 AJHSME Problems/Problem 24

## Problem

In parallelogram $ABCD$, $\overline{DE}$ is the altitude to the base $\overline{AB}$ and $\overline{DF}$ is the altitude to the base $\overline{BC}$. [Note: Both pictures represent the same parallelogram.] If $DC=12$, $EB=4$, and $DE=6$, then $DF=$ $[asy] unitsize(12); pair A,B,C,D,P,Q,W,X,Y,Z; A = (0,0); B = (12,0); C = (20,6); D = (8,6); W = (18,0); X = (30,0); Y = (38,6); Z = (26,6); draw(A--B--C--D--cycle); draw(W--X--Y--Z--cycle); P = (8,0); Q = (758/25,6/25); dot(A); dot(B); dot(C); dot(D); dot(W); dot(X); dot(Y); dot(Z); dot(P); dot(Q); draw(A--B--C--D--cycle); draw(W--X--Y--Z--cycle); draw(D--P); draw(Z--Q); label("A",A,SW); label("B",B,SE); label("C",C,NE); label("D",D,NW); label("E",P,S); label("A",W,SW); label("B",X,S); label("C",Y,NE); label("D",Z,NW); label("F",Q,E); [/asy]$ $\text{(A)}\ 6.4 \qquad \text{(B)}\ 7 \qquad \text{(C)}\ 7.2 \qquad \text{(D)}\ 8 \qquad \text{(E)}\ 10$

## Solution

Note that $\overline{DE}(\overline{AB})=\overline{DF}(\overline{BC})=Area(ABCD)$. We will try to find $BC$, $AB$, & $DE$.

First, note that $DE=6$ and $AB=CD=12$, by properties of a parallelogram. Also, $AD=BC$,.

Since $\angle DEA$ is a right angle, we can use the pythagorean theorem: $$(AE)^2+(ED)^2=(AD)^2$$ $$\sqrt{(AB-4)^2+6^2}=AD=\sqrt{8^2+36}=\sqrt{64+36}=\sqrt{100}=10$$ $$AD=BC=10$$

Now we can finally substitute: $$6(12)=DF(10)$$ $$DF=\frac{72}{10}=7.2 \Rightarrow \mathrm{(C)}$$