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Difference between revisions of "1989 AJHSME Problems/Problem 15"

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==Solution 2==
 
==Solution 2==
  
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Notice that BEDC is a trapezoid. Therefore its area is <cmath>8(\frac{6+10}{2})=8(\frac{16}{2})=8(8)=64\Rightarrow \mathrm{(D)}</cmath>
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Notice that <math>BEDC</math> is a trapezoid. Therefore its area is <cmath>8\left(\frac{6+10}{2}\right)=8\left(\frac{16}{2}\right)=8(8)=64\Rightarrow \mathrm{(D)}</cmath>
  
 
==See Also==
 
==See Also==
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{{AJHSME box|year=1989|num-b=14|num-a=16}}
 
{{AJHSME box|year=1989|num-b=14|num-a=16}}
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 23:03, 4 July 2013

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 \[8\left(\frac{6+10}{2}\right)=8\left(\frac{16}{2}\right)=8(8)=64\Rightarrow \mathrm{(D)}\]

See Also

1989 AJHSME (Problems • Answer Key • 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
All AJHSME/AMC 8 Problems and Solutions

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