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Difference between revisions of "1950 AHSME Problems/Problem 47"

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\textbf{(E)}\ x=\dfrac{1}{2}b</math>
 
\textbf{(E)}\ x=\dfrac{1}{2}b</math>
  
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==Solution 1==
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==Solution 1:Similarity==
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Draw the triangle, and note that the small triangle formed by taking away the rectangle and the two small portions left is similar to the big triangle, so the proportions of the heights is equal to the proportions of the sides. In particular, we get <math>\dfrac{2x}{b} = \dfrac{h - x}{h} \implies 2xh = bh - bx \implies (2h + b)x = bh \implies x = \dfrac{bh}{2h + b}</math>. The answer is <math>\boxed{\textbf{(C)}}</math>.
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Draw the triangle.The small triangle formed by taking away the rectangle and the two small portions left is similar to the big triangle.Hence, the proportions of the heights is equal to the proportions of the sides.  
  
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==Solution 2==
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In particular, we get
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You can also use area to solve this problem. Draw the triangle, and note that the small triangle formed by taking away the rectangle and the two small portions left is similar to the big triangle. Therefore the smaller triangle (similar to the big one) has an area of <math>\dfrac{2x(h-x)}{2}=x(h-x)</math>. Now note that because the rectangle is right, and the two other pieces are complementary triangles, we can add them together to create one triangle. Therefore the area of these two triangles is <math>\dfrac{x(b-2x)}{2}</math>. This is due to the fact that we subtract <math>2x</math> from the base of the triangle. Lastly the area of the rectangle is <math>2x^2</math>. These areas together sum to the area of the big rectangle, which is <math>\dfrac{bh}{2}</math>. Solving we get that <math>x = \dfrac{bh}{2h + b}</math>. The answer is <math>\boxed{\textbf{(C)}}</math>.
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<cmath>\dfrac{2x}{b} = \dfrac{h - x}{h} \implies 2xh = bh - bx \implies (2h + b)x = bh \implies x = \dfrac{bh}{2h + b}</cmath>
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The answer is <math>\boxed{\textbf{(C)}}</math>.
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==Solution 2: Area==
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The smaller triangle at the top has an area of <cmath>\dfrac{2x(h-x)}{2}=x(h-x)</cmath>
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Because the two other pieces are complementary triangles, we can add them together to create one triangle with area
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<cmath>\dfrac{x(b-2x)}{2}</cmath>
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Lastly the area of the rectangle is <math>2x^2</math>. These areas together sum to the area of the big rectangle, which is <math>\dfrac{bh}{2}</math>. Hence,
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<cmath>\dfrac{bh}{2}=2x^2 + \dfrac{x(b-2x)}{2}+x(h-x)</cmath>
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Solving we get that <cmath>x = \dfrac{bh}{2h + b}</cmath>
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The answer is <math>\boxed{\textbf{(C)}}</math>.
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 01:24, 11 June 2024

Problem

A rectangle inscribed in a triangle has its base coinciding with the base $b$ of the triangle. If the altitude of the triangle is $h$, and the altitude $x$ of the rectangle is half the base of the rectangle, then:

$\textbf{(A)}\ x=\dfrac{1}{2}h \qquad \textbf{(B)}\ x=\dfrac{bh}{b+h} \qquad \textbf{(C)}\ x=\dfrac{bh}{2h+b} \qquad \textbf{(D)}\ x=\sqrt{\dfrac{hb}{2}} \qquad \textbf{(E)}\ x=\dfrac{1}{2}b$

Solution 1:Similarity

Draw the triangle.The small triangle formed by taking away the rectangle and the two small portions left is similar to the big triangle.Hence, the proportions of the heights is equal to the proportions of the sides.

In particular, we get \[\dfrac{2x}{b} = \dfrac{h - x}{h} \implies 2xh = bh - bx \implies (2h + b)x = bh \implies x = \dfrac{bh}{2h + b}\] The answer is $\boxed{\textbf{(C)}}$.

Solution 2: Area

The smaller triangle at the top has an area of \[\dfrac{2x(h-x)}{2}=x(h-x)\]

Because the two other pieces are complementary triangles, we can add them together to create one triangle with area \[\dfrac{x(b-2x)}{2}\]

Lastly the area of the rectangle is $2x^2$. These areas together sum to the area of the big rectangle, which is $\dfrac{bh}{2}$. Hence, \[\dfrac{bh}{2}=2x^2 + \dfrac{x(b-2x)}{2}+x(h-x)\]

Solving we get that \[x = \dfrac{bh}{2h + b}\] The answer is $\boxed{\textbf{(C)}}$.

Video Solution

https://www.youtube.com/watch?v=l4lAvs2P_YA&t=169s

~MathProblemSolvingSkills.com


See Also

1950 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 46 		Followed by
Problem 48
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All AHSME Problems and Solutions

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