1989 AJHSME Problems/Problem 15

Problem

The area of the shaded region $\text{BEDC}$ in parallelogram $\text{ABCD}$ is

$[asy] unitsize(10); pair A,B,C,D,E; A=origin; B=(4,8); C=(14,8); D=(10,0); E=(4,0); draw(A--B--C--D--cycle); fill(B--E--D--C--cycle,gray); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,S); label("$10$",(9,8),N); label("$6$",(7,0),S); label("$8$",(4,4),W); draw((3,0)--(3,1)--(4,1)); [/asy]$

$\text{(A)}\ 24 \qquad \text{(B)}\ 48 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80$

Solution 1

Let $[ABC]$ denote the area of figure $ABC$ .

Clearly, $[BEDC]=[ABCD]-[ABE]$ . Using basic area formulas,

$[ABCD]=(BC)(BE)=80$ $[ABE]=(BE)(AE)/2 = 4(AE)$

Since $AE+ED=BC=10$ and $ED=6$ , $AE=4$ and the area of $\triangle ABE$ is $4(4)=16$ .

Finally, we have $[BEDC]=80-16=64\rightarrow \boxed{\text{D}}$

Solution 2

Notice that $BEDC$ is a trapezoid. Therefore its area is $8\left(\frac{6+10}{2}\right)=8\left(\frac{16}{2}\right)=8(8)=64\Rightarrow \mathrm{(D)}$

1989 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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1989 AJHSME Problems/Problem 15

Contents

Problem

Solution 1

Solution 2

See Also