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Difference between revisions of "1985 AHSME Problems/Problem 17"

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==Solution==
 
==Solution==
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Let <math> E </math> be the intersection of line <math> L </math> and <math> \stackrel{\longleftrightarrow}{BD} </math>. Because <math> AE </math> is the altitude to the hypotenuse of right triangle <math> ABD </math>, we have <math>(AE)^2=BE \cdot ED</math>. Thus, <math> AE^2=(1)(2)\implies AE=\sqrt{2} </math>. Now we use <math> A=\frac{1}{2}bh </math> on <math> \triangle ABD </math> to get <math> [ABD]=\frac{1}{2}(3)(\sqrt{2})=\frac{3\sqrt{2}}{2} </math>. Now we have to double it to get the area of the entire rectangle: <math> 2\left(\frac{3\sqrt{2}}{2}\right)=3\sqrt{2}\approx4.2, \boxed{\text{B}} </math>.
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Let <math> E </math> be the intersection of line <math> L </math> and <math> \stackrel{\longleftrightarrow}{BD} </math>. Because <math> AE </math> is the altitude to the hypotenuse of right triangle <math> ABD </math>, we have <math>(AE)^2=BE \cdot ED</math>. Thus, <math> AE^2=(1)(2)\implies AE=\sqrt{2} </math>. Now we use <math> A=\frac{1}{2}bh </math> on <math> \triangle ABD </math> to get <math> [ABD]=\frac{1}{2}(3)(\sqrt{2})=\frac{3\sqrt{2}}{2} </math>. Now we have to double it to get the area of the entire rectangle: <math> 2\left(\frac{3\sqrt{2}}{2}\right)=3\sqrt{2}\approx4.2,</math> Which leads to the answer <math>\boxed{\text{(B) 4.2}} </math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=16|num-a=18}}
 
{{AHSME box|year=1985|num-b=16|num-a=18}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 07:39, 20 November 2023

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 intersection of line $L$ and $\stackrel{\longleftrightarrow}{BD}$. Because $AE$ is the altitude to the hypotenuse of right triangle $ABD$, we have $(AE)^2=BE \cdot ED$. Thus, $AE^2=(1)(2)\implies AE=\sqrt{2}$. Now we use $A=\frac{1}{2}bh$ on $\triangle ABD$ to get $[ABD]=\frac{1}{2}(3)(\sqrt{2})=\frac{3\sqrt{2}}{2}$. Now we have to double it to get the area of the entire rectangle: $2\left(\frac{3\sqrt{2}}{2}\right)=3\sqrt{2}\approx4.2,$ Which leads to the answer $\boxed{\text{(B) 4.2}}$.

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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