1985 AHSME Problems/Problem 17
Problem
Diagonal 
of rectangle 
is divided into three segments of length 
by parallel lines 
and 
that pass through 
and 
and are perpendicular to . The area of , 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]](http://latex.artofproblemsolving.com/e/4/3/e43f7393003b1ebc502efe5c8e1abdd4d2e253be.png)
Solution
Let 
be the intersection of line and 
. Because 
is the altitude to the hypotenuse of right triangle 
, we have 
. Thus, 
. Now we use 
on 
to get ![$[ABD]=\frac{1}{2}(3)(\sqrt{2})=\frac{3\sqrt{2}}{2}$](http://latex.artofproblemsolving.com/0/8/5/0855845d52388014e4bb091c359fd5e5b5b35bd8.png)
. Now we have to double it to get the area of the entire rectangle: 
.
See Also
| 1985 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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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