Difference between revisions of "1960 AHSME Problems/Problem 37"

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\textbf{(E)}\ x(h-x)    </math>
 
\textbf{(E)}\ x(h-x)    </math>
  
==Solution (WIP)==
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==Solution==
 
Let <math>AB=b</math>, <math>DE=h</math>, and <math>WX = YZ = x</math>.
 
Let <math>AB=b</math>, <math>DE=h</math>, and <math>WX = YZ = x</math>.
 
<asy>
 
<asy>
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Since <math>CD</math> is perpendicular to <math>AB</math>, <math>ND = WX</math>.  That means <math>CN = h-x</math>.
 
Since <math>CD</math> is perpendicular to <math>AB</math>, <math>ND = WX</math>.  That means <math>CN = h-x</math>.
 
The sides of the rectangle are parallel, so <math>XY \parallel WZ</math>.  That means by AA Similarity, <math>\triangle CXY \sim \triangle CAB</math>.  Letting <math>n</math> be the length of the base of the rectangle, that means
 
The sides of the rectangle are parallel, so <math>XY \parallel WZ</math>.  That means by AA Similarity, <math>\triangle CXY \sim \triangle CAB</math>.  Letting <math>n</math> be the length of the base of the rectangle, that means
<cmath>\frac{h-x}{n} = \frac{}{}</cmath>
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<cmath>\frac{h-x}{n} = \frac{h}{b}</cmath>
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<cmath>n = \frac{b(h-x)}{h}</cmath>
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Thus, the area of the rectangle is <math>\frac{bx}{h}(h-x)</math>, which is answer choice <math>\boxed{\textbf{(A)}}</math>.
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==Video Solution==
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https://youtu.be/ZdM2ou5Gsuw?t=172
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~MathProblemSolvingSkills.com
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==See Also==
 
==See Also==
 
{{AHSME 40p box|year=1960|num-b=36|num-a=38}}
 
{{AHSME 40p box|year=1960|num-b=36|num-a=38}}
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[[Category:Intermediate Geometry Problems]]

Latest revision as of 21:26, 28 December 2023

Problem

The base of a triangle is of length $b$, and the altitude is of length $h$. A rectangle of height $x$ is inscribed in the triangle with the base of the rectangle in the base of the triangle. The area of the rectangle is:

$\textbf{(A)}\ \frac{bx}{h}(h-x)\qquad \textbf{(B)}\ \frac{hx}{b}(b-x)\qquad \textbf{(C)}\ \frac{bx}{h}(h-2x)\qquad \textbf{(D)}\ x(b-x)\qquad \textbf{(E)}\ x(h-x)$

Solution

Let $AB=b$, $DE=h$, and $WX = YZ = x$. [asy] pair A=(0,0),B=(56,0),C=(20,48),D=(20,0),W=(10,0),X=(10,24),Y=(38,24),Z=(38,0); draw(A--B--C--A); draw((10,0)--(10,24)--(38,24)--(38,0)); draw(C--D); dot(A); dot(B); dot(C); dot(D); dot(W); dot(X); dot(Y); dot(Z); dot((20,24)); label("$A$",A,S); label("$B$",B,S); label("$C$",C,N); label("$D$",D,S); label("$W$",W,S); label("$X$",X,NW); label("$Y$",Y,NE); label("$Z$",Z,S); label("$N$",(20,24),NW); [/asy] Since $CD$ is perpendicular to $AB$, $ND = WX$. That means $CN = h-x$. The sides of the rectangle are parallel, so $XY \parallel WZ$. That means by AA Similarity, $\triangle CXY \sim \triangle CAB$. Letting $n$ be the length of the base of the rectangle, that means \[\frac{h-x}{n} = \frac{h}{b}\] \[n = \frac{b(h-x)}{h}\] Thus, the area of the rectangle is $\frac{bx}{h}(h-x)$, which is answer choice $\boxed{\textbf{(A)}}$.


Video Solution

https://youtu.be/ZdM2ou5Gsuw?t=172

~MathProblemSolvingSkills.com


See Also

1960 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 36
Followed by
Problem 38
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All AHSME Problems and Solutions