Difference between revisions of "1993 AJHSME Problems/Problem 18"
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<math>\text{(A)}\ 320 \qquad \text{(B)}\ 325 \qquad \text{(C)}\ 330 \qquad \text{(D)}\ 335 \qquad \text{(E)}\ 340</math> | <math>\text{(A)}\ 320 \qquad \text{(B)}\ 325 \qquad \text{(C)}\ 330 \qquad \text{(D)}\ 335 \qquad \text{(E)}\ 340</math> | ||
| + | |||
| + | ==Solution== | ||
| + | The area of the quadrilateral <math>ABDF</math> is equal to the areas of the two right triangles <math>\triangle BCD</math> and <math>\triangle EFD</math> subtracted from the area of the rectangle <math>ABCD</math>. Because <math>B</math> and <math>F</math> 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> | ||
| + | |||
| + | <cmath>(20)(32)-\frac{(16)(20)}{2}-\frac{(10)(32)}{2} = 640-160-160 = \boxed{\text{(A)}\ 320}</cmath> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AJHSME box|year=1993|num-b=17|num-a=19}} | ||
| + | {{MAA Notice}} | ||
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. 
