1989 AJHSME Problems/Problem 15
Contents
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 1
Let ![$[ABC]$](http://latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
denote the area of figure 
.
Clearly, ![$[BEDC]=[ABCD]-[ABE]$](http://latex.artofproblemsolving.com/0/0/6/006d04a81a75b38705d476bffb303d3b061515bd.png)
. 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 ![$[BEDC]=80-16=64\rightarrow \boxed{\text{D}}$](http://latex.artofproblemsolving.com/2/4/4/2446cf505d3655172dea4cfbf3b30697e7492aa7.png)
Solution 2
Notice that BEDC is a trapezoid. Therefore its area is ![\[8(\frac{6+10}{2})=8(\frac{16}{2})=8(8)=64\Rightarrow \mathrm{(D)}\]](http://latex.artofproblemsolving.com/b/1/6/b16ab62dd95868640ea7147f5e3cbc90ed679bfe.png)
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 | ||
