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 26"

(Created page with "==Problem 26== The base of a triangle is <math>15</math> inches. Two lines are drawn parallel to the base, terminating in the other two sides, and dividing the triangle into...")
 
(→‎Problem 26)
 
(2 intermediate revisions by 2 users not shown)
Line 7: Line 7:
 
\textbf{(C)}\ 4\sqrt{3}\text{ inches}\qquad
 
\textbf{(C)}\ 4\sqrt{3}\text{ inches}\qquad
 
\textbf{(D)}\ 7.5\text{ inches}\\  
 
\textbf{(D)}\ 7.5\text{ inches}\\  
−
\textbf{(E)}\ \text{none of these}    </math>
+
\textbf{(4)}\ \text{none of these}    </math>
  
 
==Solution==
 
==Solution==
Line 15: Line 15:
  
 
Hence our answer is <math>\textbf{(A)}</math>
 
Hence our answer is <math>\textbf{(A)}</math>
 +
 +
==See Also==
 +
{{AHSME 50p box|year=1953|num-b=25|num-a=27}}
 +
 +
{{MAA Notice}}

Latest revision as of 21:51, 24 February 2020

Problem 26

The base of a triangle is $15$ inches. Two lines are drawn parallel to the base, terminating in the other two sides, and dividing the triangle into three equal areas. The length of the parallel closer to the base is:

$\textbf{(A)}\ 5\sqrt{6}\text{ inches} \qquad \textbf{(B)}\ 10\text{ inches} \qquad \textbf{(C)}\ 4\sqrt{3}\text{ inches}\qquad \textbf{(D)}\ 7.5\text{ inches}\\ \textbf{(4)}\ \text{none of these}$

Solution

Let the triangle be $\triangle ABC$ where $BC$ is the base. Then let the parallels be $MN$ and $PQ$, where $PQ$ is closer to our base $BC$.

It's obvious that $\triangle APQ \sim \triangle ABC$, where $|\triangle APQ|:|\triangle ABC|=2:3$, so $\frac{PQ}{BC}=\sqrt{\frac{2}{3}}$. Since we know $BC=15$, $PQ=\frac{15\sqrt{2}}{\sqrt{3}}=5\sqrt{6}$

Hence our answer is $\textbf{(A)}$

See Also

1953 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 25 		Followed by
Problem 27
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_26&oldid=118353"