Difference between revisions of "1972 AHSME Problems/Problem 25"

(Created page with "==Solution== We note that <math>25^2+60^2=65^2</math> and <math>39^2+52^2=65^2</math> so our answer is <math>\boxed{65}</math>. -Pleaseletmewin")
 
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Inscribed in a circle is a quadrilateral having sides of lengths <math>25,~39,~52</math>, and <math>60</math> taken consecutively. The diameter of this circle has length
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<math>\textbf{(A) }62\qquad \textbf{(B) }63\qquad \textbf{(C) }65\qquad \textbf{(D) }66\qquad  \textbf{(E) }69</math>
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==Solution==
 
==Solution==
  
We note that <math>25^2+60^2=65^2</math> and <math>39^2+52^2=65^2</math> so our answer is <math>\boxed{65}</math>.
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We note that <math>25^2+60^2=65^2</math> and <math>39^2+52^2=65^2</math> so our answer is <math>\boxed{C}</math>.
  
 
-Pleaseletmewin
 
-Pleaseletmewin

Revision as of 16:02, 2 August 2020

Inscribed in a circle is a quadrilateral having sides of lengths $25,~39,~52$, and $60$ taken consecutively. The diameter of this circle has length

$\textbf{(A) }62\qquad \textbf{(B) }63\qquad \textbf{(C) }65\qquad \textbf{(D) }66\qquad  \textbf{(E) }69$

Solution

We note that $25^2+60^2=65^2$ and $39^2+52^2=65^2$ so our answer is $\boxed{C}$.

-Pleaseletmewin

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