Difference between revisions of "1989 AJHSME Problems/Problem 15"
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==Solution== | ==Solution== | ||
− | Let <math> | + | Let <math>[ABC]</math> denote the area of figure <math>ABC</math>. |
− | Clearly, <math> | + | Clearly, <math>[BEDC]=[ABCD]-[ABE]</math>. Using basic area formulas, |
− | < | + | |
− | < | + | <center><math>[ABCD]=(BC)(BE)=80</math></center> |
+ | |||
+ | <center><math>[ABE]=(BE)(AE)/2 = 4(AE)</math></center> | ||
Since <math>AE+ED=BC=10</math> and <math>ED=6</math>, <math>AE=4</math> and the area of <math>\triangle ABE</math> is <math>4(4)=16</math>. | Since <math>AE+ED=BC=10</math> and <math>ED=6</math>, <math>AE=4</math> and the area of <math>\triangle ABE</math> is <math>4(4)=16</math>. | ||
− | Finally, we have <math> | + | Finally, we have <math>[BEDC]=80-16=64\rightarrow \boxed{\text{D}}</math> |
==See Also== | ==See Also== |
Revision as of 21:23, 25 April 2010
Problem
The area of the shaded region in parallelogram is
Solution
Let denote the area of figure .
Clearly, . Using basic area formulas,
Since and , and the area of is .
Finally, we have
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 |