Difference between revisions of "1989 AJHSME Problems/Problem 15"
5849206328x (talk | contribs) (New page: ==Problem== The area of the shaded region <math>\text{BEDC}</math> in parallelogram <math>\text{ABCD}</math> is <asy> unitsize(10); pair A,B,C,D,E; A=origin; B=(4,8); C=(14,8); D...) |
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<cmath>!ABE!=(BE)(AE)/2 = 4(AE)</cmath> | <cmath>!ABE!=(BE)(AE)/2 = 4(AE)</cmath> | ||
| − | Since <math>AE+ | + | 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>!BEDC!=80-16=64\rightarrow \boxed{\text{D}}</math> | Finally, we have <math>!BEDC!=80-16=64\rightarrow \boxed{\text{D}}</math> | ||
Revision as of 05:46, 25 April 2010
Problem
The area of the shaded region 
in parallelogram 
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]](http://latex.artofproblemsolving.com/6/0/9/6092322b7c3f4d13fc5df8d5bf92420b2010448d.png)
Solution
Let 
denote the area of figure 
.
Clearly, 
. Using basic area formulas,
![\[!ABCD!=(BC)(BE)=80\]](http://latex.artofproblemsolving.com/e/1/9/e199b403c3564d8429411cd53acca719d5e6ce20.png)
![\[!ABE!=(BE)(AE)/2 = 4(AE)\]](http://latex.artofproblemsolving.com/6/b/5/6b5f453767aaef6e9e115a5e4461ed61b213d02c.png)
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 | ||
