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Difference between revisions of "1998 AHSME Problems/Problem 8"
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The area of the trapezoid is <math>\frac{1}{3}</math>, and the shorter base and height are both <math>\frac{1}{2}</math>.  Therefore, <cmath>\frac{1}{3}=\frac{1}{2}\cdot \frac{1}{2}\cdot \left(\frac{1}{2}+x\right) \Rightarrow x=\frac{5}{6}\rightarrow \boxed{\text{D}}</cmath>
 
The area of the trapezoid is <math>\frac{1}{3}</math>, and the shorter base and height are both <math>\frac{1}{2}</math>.  Therefore, <cmath>\frac{1}{3}=\frac{1}{2}\cdot \frac{1}{2}\cdot \left(\frac{1}{2}+x\right) \Rightarrow x=\frac{5}{6}\rightarrow \boxed{\text{D}}</cmath>
 +
 +
== Solution 3 ==
 +
 +
Divide the pentagon into 2 small congruent trapezoids by extending the common shorter base of the 2 larger trapezoids.
 +
 +
Since each of the smaller trapezoids has its area half each of the larger trapezoids, and each of them has a base <math>\frac{1}{2}</math>, we have
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<cmath>b_{large}+\frac{1}{2}=2(b_{small}+ \frac{1}{2})</cmath>
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<cmath>x+\frac{1}{2}=2((1-x)+\frac{1}{2})</cmath>
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<cmath>x=\frac{5}{6}\boxed{D}</cmath>
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 +
~ Nafer
  
 
== See also ==
 
== See also ==

Latest revision as of 10:28, 2 December 2019

Problem

A square with sides of length $1$ is divided into two congruent trapezoids and a pentagon, which have equal areas, by joining the center of the square with points on three of the sides, as shown. Find $x$, the length of the longer parallel side of each trapezoid.

[asy] pointpen = black; pathpen = black; D(unitsquare); D((0,0)); D((1,0)); D((1,1)); D((0,1)); D(D((.5,.5))--D((1,.5))); D(D((.17,1))--(.5,.5)--D((.17,0))); MP("x",(.58,1),N); [/asy]

$\mathrm{(A) \ } \frac 35 \qquad \mathrm{(B) \ } \frac 23 \qquad \mathrm{(C) \ } \frac 34 \qquad \mathrm{(D) \ } \frac 56 \qquad \mathrm{(E) \ } \frac 78$

Solution

Solution 1

[asy] pointpen = black; pathpen = black; D(unitsquare); D((0,0)); D((1,0)); D((1,1)); D((0,1)); D(D((.5,.5))--D((1,.5))); D(D((.17,1))--(.5,.5)--D((.5,1)));D(D((1-.17,1))--(.5,.5)--D((.17,0))); D((.17,1)--(.17,0));D((1-.17,1)--(1-.17,.5));D((0,.5)--(.5,.5)); MP("x",(.58,1),N); MP("I",(.17/2,.25),(0,0));MP("I",(.17/2,.75),(0,0));MP("I",(1-.17/2,.75),(0,0));MP("II",(.5-.17,.4),(0,0));MP("II",(.5-.17,.6),(0,0));MP("II",(.5-.17,.9),(0,0));MP("II",(.5+.17,.9),(0,0));MP("II",(.5+.17,.6),(0,0)); [/asy]

Then $2[I]+2[II] = [I]+3[II] \Longrightarrow [I]=[II]$. Let the shorter side of $I$ be $m$ and the base of $II$ be $n$ such that $m+2n = x$; then $[I]=[II]$ implies that $2m=n$, and since $2m + 2n = 1$ it follows that $m = \frac 16$ and $x = \frac 56 \Longrightarrow \mathbf{(D)}$.

Solution 2

The area of the trapezoid is $\frac{1}{3}$, and the shorter base and height are both $\frac{1}{2}$. Therefore, \[\frac{1}{3}=\frac{1}{2}\cdot \frac{1}{2}\cdot \left(\frac{1}{2}+x\right) \Rightarrow x=\frac{5}{6}\rightarrow \boxed{\text{D}}\]

Solution 3

Divide the pentagon into 2 small congruent trapezoids by extending the common shorter base of the 2 larger trapezoids.

Since each of the smaller trapezoids has its area half each of the larger trapezoids, and each of them has a base $\frac{1}{2}$, we have \[b_{large}+\frac{1}{2}=2(b_{small}+ \frac{1}{2})\] \[x+\frac{1}{2}=2((1-x)+\frac{1}{2})\] \[x=\frac{5}{6}\boxed{D}\]

~ Nafer

See also

1998 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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