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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<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> | + | 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> | |
| − | Finally, we have <math> | + | <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>. | ||
| + | |||
| + | Finally, we have <math>[BEDC]=80-16=64\rightarrow \boxed{\text{D}}</math> | ||
| + | |||
| + | ==Solution 2== | ||
| + | |||
| + | Notice that <math>BEDC</math> 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== | ||
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{{AJHSME box|year=1989|num-b=14|num-a=16}} | {{AJHSME box|year=1989|num-b=14|num-a=16}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 23:03, 4 July 2013
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 
is a trapezoid. Therefore its area is ![\[8\left(\frac{6+10}{2}\right)=8\left(\frac{16}{2}\right)=8(8)=64\Rightarrow \mathrm{(D)}\]](http://latex.artofproblemsolving.com/2/2/1/221ad895a6fdb1c46904516cf923087b2f9872d3.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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
