Difference between revisions of "1953 AHSME Problems/Problem 13"
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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]] | ||
+ | {{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:
Solution
We know that the area of a trapezoid is where is the median and the area of a triangle is . We know that and the height of the trapezoid is congruent to the height of the triangle. Thus, we get , so .
See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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All AHSME Problems and Solutions |
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