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 "1958 AHSME Problems/Problem 36"

(Created page with "== Problem == The sides of a triangle are <math> 30</math>, <math> 70</math>, and <math> 80</math> units. If an altitude is dropped upon the side of length <math> 80</math>, the ...")
 
m (→‎Solution)
 
(3 intermediate revisions by the same user not shown)
Line 9: Line 9:
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{}</math>
+
Let the shorter segment be <math>x</math> and the altitude be <math>y</math>. The larger segment is then <math>80-x</math>.  By the [[Pythagorean Theorem]]
 +
, <cmath>30^2-y^2=x^2 \qquad(1)</cmath>
 +
and <cmath>(80-x)^2=70^2-y^2 \qquad(2)</cmath>
 +
Adding <math>(1)</math> and <math>(2)</math> and simplifying gives <math>x=15</math>. Therefore, the answer is <math>80-15=\boxed{\textbf{(D)}~65}</math>
 +
 
 +
~megaboy6679
  
 
== See Also ==
 
== See Also ==

Latest revision as of 01:15, 29 January 2023

Problem

The sides of a triangle are $30$, $70$, and $80$ units. If an altitude is dropped upon the side of length $80$, the larger segment cut off on this side is:

$\textbf{(A)}\ 62\qquad \textbf{(B)}\ 63\qquad \textbf{(C)}\ 64\qquad \textbf{(D)}\ 65\qquad \textbf{(E)}\ 66$

Solution

Let the shorter segment be $x$ and the altitude be $y$. The larger segment is then $80-x$. By the Pythagorean Theorem , \[30^2-y^2=x^2 \qquad(1)\] and \[(80-x)^2=70^2-y^2 \qquad(2)\] Adding $(1)$ and $(2)$ and simplifying gives $x=15$. Therefore, the answer is $80-15=\boxed{\textbf{(D)}~65}$

~megaboy6679

See Also

1958 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 35 		Followed by
Problem 37
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=1958_AHSME_Problems/Problem_36&oldid=188293"