Difference between revisions of "1989 AJHSME Problems/Problem 15"
(→Solution) |
|||
(4 intermediate revisions by 2 users not shown) | |||
Line 16: | Line 16: | ||
<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> |
+ | |||
+ | <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> |
+ | |||
+ | ==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== | ||
Line 32: | Line 38: | ||
{{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 00:03, 5 July 2013
Contents
Problem
The area of the shaded region in parallelogram
is
Solution 1
Let denote the area of figure
.
Clearly, . 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
Solution 2
Notice that 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.