# 1985 AHSME Problems/Problem 17

## Problem

Diagonal $DB$ of rectangle $ABCD$ is divided into three segments of length $1$ by parallel lines $L$ and $L'$ that pass through $A$ and $C$ and are perpendicular to $DB$. The area of $ABCD$, rounded to the one decimal place, is

$[asy] defaultpen(linewidth(0.7)+fontsize(10)); real x=sqrt(6), y=sqrt(3), a=0.4; pair D=origin, A=(0,y), B=(x,y), C=(x,0), E=foot(C,B,D), F=foot(A,B,D); real r=degrees(B); pair M1=F+3*dir(r)*dir(90), M2=F+3*dir(r)*dir(-90), N1=E+3*dir(r)*dir(90), N2=E+3*dir(r)*dir(-90); markscalefactor=0.02; draw(B--C--D--A--B--D^^M1--M2^^N1--N2^^rightanglemark(A,F,B)^^rightanglemark(N1,E,B)); pair W=A+a*dir(135), X=B+a*dir(45), Y=C+a*dir(-45), Z=D+a*dir(-135); label("A", A, NE); label("B", B, NE); label("C", C, dir(0)); label("D", D, dir(180)); label("L", (x/2,0), SW); label("L^\prime", C, SW); label("1", D--F, NW); label("1", F--E, SE); label("1", E--B, SE); clip(W--X--Y--Z--cycle);[/asy]$

$\mathrm{(A)\ } 4.1 \qquad \mathrm{(B) \ }4.2 \qquad \mathrm{(C) \ } 4.3 \qquad \mathrm{(D) \ } 4.4 \qquad \mathrm{(E) \ }4.5$

## Solution

Let $E$ be the point of intersection of $L$ and $\overline{BD}$. Then, because $AE$ is the altitude to the hypotenuse of right triangle $ABD$, triangles $ADE$ and $BAE$ are similar, giving $$\frac{AE}{BE} = \frac{ED}{EA},$$ and so \begin{align*}AE &= \sqrt{BE \cdot ED} \\ &= \sqrt{(1+1)(1)} \\ &= \sqrt{2}.\end{align*} Thus, taking $BD$ and $AE$ as the base and perpendicular height, respectively, of triangle $ABD$, we may compute its area as $\frac{1}{2}(3)\left(\sqrt{2}\right) = \frac{3\sqrt{2}}{2}$. By symmetry, the area of the entire rectangle $ABCD$ is $$2\left(\frac{3\sqrt{2}}{2}\right) = 3\sqrt{2} \approx (3)(1.4) = \boxed{\text{(B)} \ 4.2}.$$