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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 ...")
 
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== Solution ==
 
== Solution ==
−
<math>\fbox{}</math>
+
Let the shorter segment be x. The larger segment is therefore 80-x. Let the altitude be y. 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{(D)~65}</math>
  
 +
~megaboy6679
 
== See Also ==
 
== See Also ==
  
 
{{AHSME 50p box|year=1958|num-b=35|num-a=37}}
 
{{AHSME 50p box|year=1958|num-b=35|num-a=37}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:19, 27 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. The larger segment is therefore 80-x. Let the altitude be y. 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 $80-15=\boxed{(D)~65}$

~megaboy6679

See Also

1958 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 35 		Followed by
Problem 37
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All AHSME Problems and Solutions

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