Difference between revisions of "1989 AJHSME Problems/Problem 15"
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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== | + | ==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> | ||
+ | |||
+ | ==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> | ||
==See Also== | ==See Also== |
Revision as of 19:23, 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,
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 |