Difference between revisions of "1993 AJHSME Problems/Problem 18"
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Latest revision as of 00:05, 11 November 2019
Problem
The rectangle shown has length 
, width 
, and 
and 
are midpoints of 
and 
, respectively. The area of quadrilateral 
is
![[asy] pair A,B,C,D,EE,F; A = (0,20); B = (16,20); C = (32,20); D = (32,0); EE = (0,0); F = (0,10); draw(A--C--D--EE--cycle); draw(B--D--F); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); label("$A$",A,NW); label("$B$",B,N); label("$C$",C,NE); label("$D$",D,SE); label("$E$",EE,SW); label("$F$",F,W); [/asy]](http://latex.artofproblemsolving.com/2/8/3/283e16cb4390b4628de8a0c51335af0769976cd0.png)
Solution
The area of the quadrilateral is equal to the areas of the two right triangles 
and 
subtracted from the area of the rectangle 
. Because and are midpoints, we know the dimensions of the two right triangles.
![[asy] pair A,B,C,D,EE,F; A = (0,20); B = (16,20); C = (32,20); D = (32,0); EE = (0,0); F = (0,10); draw(A--C--D--EE--cycle); draw(B--D--F); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); label("$A$",A,NW); label("$B$",B,N); label("$C$",C,NE); label("$D$",D,SE); label("$E$",EE,SW); label("$F$",F,W); label("$16$",A--B,N); label("$16$",B--C,N); label("$32$",E--D,S); label("$10$",E--F,W); label("$10$",A--F,W); label("$20$",C--D,E); [/asy]](http://latex.artofproblemsolving.com/f/2/d/f2d859b431568b215aca56f2115eba716c543604.png)
![\[(20)(32)-\frac{(16)(20)}{2}-\frac{(10)(32)}{2} = 640-160-160 = \boxed{\text{(A)}\ 320}\]](http://latex.artofproblemsolving.com/e/5/7/e57bc1cf55eae8a0148d9db1c9b0c1635393a230.png)
See Also
| 1993 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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. 
