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1989 AJHSME Problems/Problem 15

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Problem

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

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

Solution

Let $!ABC!$ denote the area of figure $ABC$ .

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

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

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

See Also

1989 AJHSME (Problems • Resources)
Preceded by Problem 14	Followed by Problem 16
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

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Category: Introductory Geometry Problems

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