Difference between revisions of "1998 AHSME Problems/Problem 8"

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

@@ Line 23: / Line 23: @@
 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
+<cmath>b_{large}+\frac{1}{2}=2(b_{small}+ \frac{1}{2})</cmath>
+<cmath>x+\frac{1}{2}=2((1-x)+\frac{1}{2})</cmath>
+<cmath>x=\frac{5}{6}\boxed{D}</cmath>
+~ Nafer
 == See also ==
-{{AHSME box|year=1998|num-b=26|num-a=28}}
+{{AHSME box|year=1998|num-b=7|num-a=9}}
 [[Category:Introductory Geometry Problems]]
+{{MAA Notice}}

1998 AHSME (Problems • Answer Key • Resources)
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