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 $\angle BAD$ 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: $6(12)=DF(10)$ $DF=\frac{72}{10}=7.2 \Rightarrow \mathrm{(C)}$

Page

Toolbox

Search

1995 AJHSME Problems/Problem 24

Problem

Solution