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

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(Solution)
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<math>\text{(A)}\ 24 \qquad \text{(B)}\ 48 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80</math>
 
<math>\text{(A)}\ 24 \qquad \text{(B)}\ 48 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 64 \qquad \text{(E)}\ 80</math>
  
==Solution==
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==Solution 1==
  
 
Let <math>[ABC]</math> denote the area of figure <math>ABC</math>.
 
Let <math>[ABC]</math> denote the area of figure <math>ABC</math>.
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Finally, we have <math>[BEDC]=80-16=64\rightarrow \boxed{\text{D}}</math>
 
Finally, we have <math>[BEDC]=80-16=64\rightarrow \boxed{\text{D}}</math>
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 +
==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>
  
 
==See Also==
 
==See Also==

Revision as of 19:23, 3 June 2012

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

See Also

1989 AJHSME (ProblemsAnswer KeyResources)
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