1998 AHSME Problems/Problem 8

From AoPSWiki

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.

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...

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

Solution

Solution 1

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, $\frac{1}{3}=\frac{1}{2}\cdot \frac{1}{2}\cdot \left(\frac{1}{2}+x\right) \Rightarrow x=\frac{5}{6}\rightarrow \boxed{\text{D}...$

1998 AHSME (Problems)
Preceded by Problem 26	Followed by Problem 28
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

Personal tools

Navigation

Search

Toolbox