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Difference between revisions of "1958 AHSME Problems/Problem 36"

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m (→‎Solution)
 
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== Solution ==
 
== Solution ==
 
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]]
 
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>
+
, <cmath>30^2-y^2=x^2 \qquad(1)</cmath>
−
and <cmath>(80-x)^2=70^2+y^2 \qquad(2)</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>
 
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
 
~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}}

Latest revision as of 00:15, 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
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All AHSME Problems and Solutions

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