Difference between revisions of "1953 AHSME Problems/Problem 13"

(Created page with "== Problem == A triangle and a trapezoid are equal in area. They also have the same altitude. If the base of the triangle is 18 inches, the median of the trapezoid is: <mat...")
 
 
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We know that the area of a trapezoid is <math>mh</math> where <math>m</math> is the median and the area of a triangle is <math>\frac{bh}{2}</math>. We know that <math>b=18</math> and the height of the trapezoid is congruent to the height of the triangle. Thus, we get <math>\frac{18h}{2}=9h=mh</math>, so <math>m=\textbf{(B) }9\text{ inches}</math>.
 
We know that the area of a trapezoid is <math>mh</math> where <math>m</math> is the median and the area of a triangle is <math>\frac{bh}{2}</math>. We know that <math>b=18</math> and the height of the trapezoid is congruent to the height of the triangle. Thus, we get <math>\frac{18h}{2}=9h=mh</math>, so <math>m=\textbf{(B) }9\text{ inches}</math>.
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==See Also==
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{{AHSME 50p box|year=1953|num-b=12|num-a=14}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 20:56, 1 April 2017

Problem

A triangle and a trapezoid are equal in area. They also have the same altitude. If the base of the triangle is 18 inches, the median of the trapezoid is:

$\textbf{(A)}\ 36\text{ inches} \qquad \textbf{(B)}\ 9\text{ inches} \qquad \textbf{(C)}\ 18\text{ inches}\\  \textbf{(D)}\ \text{not obtainable from these data}\qquad \textbf{(E)}\ \text{none of these}$

Solution

We know that the area of a trapezoid is $mh$ where $m$ is the median and the area of a triangle is $\frac{bh}{2}$. We know that $b=18$ and the height of the trapezoid is congruent to the height of the triangle. Thus, we get $\frac{18h}{2}=9h=mh$, so $m=\textbf{(B) }9\text{ inches}$.


See Also

1953 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AHSME Problems and Solutions

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