Difference between revisions of "1993 AJHSME Problems/Problem 18"

(Created page with "==Problem== The rectangle shown has length <math>AC=32</math>, width <math>AE=20</math>, and <math>B</math> and <math>F</math> are midpoints of <math>\overline{AC}</math> and <m...")
 
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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>
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==Solution==
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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.
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<asy>
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pair A,B,C,D,EE,F;
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A = (0,20); B = (16,20); C = (32,20); D = (32,0); EE = (0,0); F = (0,10);
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draw(A--C--D--EE--cycle);
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draw(B--D--F);
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dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F);
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label("$A$",A,NW);
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label("$B$",B,N);
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label("$C$",C,NE);
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label("$D$",D,SE);
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label("$E$",EE,SW);
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label("$F$",F,W);
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label("$16$",A--B,N);
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label("$16$",B--C,N);
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label("$32$",E--D,S);
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label("$10$",E--F,W);
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label("$10$",A--F,W);
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label("$20$",C--D,E);
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</asy>
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<cmath>(20)(32)-\frac{(16)(20)}{2}-\frac{(10)(32)}{2} = 640-160-160 = \boxed{\text{(A)}\ 320}</cmath>
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==See Also==
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{{AJHSME box|year=1993|num-b=17|num-a=19}}
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{{MAA Notice}}

Latest revision as of 00:05, 11 November 2019

Problem

The rectangle shown has length $AC=32$, width $AE=20$, and $B$ and $F$ are midpoints of $\overline{AC}$ and $\overline{AE}$, respectively. The area of quadrilateral $ABDF$ 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]

$\text{(A)}\ 320 \qquad \text{(B)}\ 325 \qquad \text{(C)}\ 330 \qquad \text{(D)}\ 335 \qquad \text{(E)}\ 340$

Solution

The area of the quadrilateral $ABDF$ is equal to the areas of the two right triangles $\triangle BCD$ and $\triangle EFD$ subtracted from the area of the rectangle $ABCD$. Because $B$ and $F$ 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]

\[(20)(32)-\frac{(16)(20)}{2}-\frac{(10)(32)}{2} = 640-160-160 = \boxed{\text{(A)}\ 320}\]

See Also

1993 AJHSME (ProblemsAnswer KeyResources)
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

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