Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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...")
 
 
Line 12: Line 12:
  
 
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>.
 +
 +
 +
 +
==See Also==
 +
 +
{{AHSME 50p box|year=1953|num-b=12|num-a=14}}
 +
 +
[[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:

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

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 (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_13&oldid=85126"