Difference between revisions of "1989 AJHSME Problems/Problem 15"
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==Solution 2== | ==Solution 2== | ||
− | 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> | + | Notice that BEDC 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== |
Revision as of 20:23, 3 June 2012
Contents
Problem
The area of the shaded region in parallelogram
is
Solution 1
Let denote the area of figure
.
Clearly, . Using basic area formulas,
![$[ABCD]=(BC)(BE)=80$](http://latex.artofproblemsolving.com/3/0/b/30bf642b7eab0db2f21d65aaf49ebc74ab071cf2.png)
![$[ABE]=(BE)(AE)/2 = 4(AE)$](http://latex.artofproblemsolving.com/3/8/c/38c5d7aa2f721d91ac139577a5fa12950ce4f86d.png)
Since and
,
and the area of
is
.
Finally, we have
Solution 2
Notice that BEDC is a trapezoid. Therefore its area is
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 |