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A square and an equilateral triangle have the same perimeter. Let Abe the area of the circle circumscribed about the square and B be the area of the circle circumscribed about the triangle. Find A/B

Mellie  Nov 20, 2015

Best Answer 

 #1
avatar+91038 
+15

A square and an equilateral triangle have the same perimeter. Let Abe the area of the circle circumscribed about the square and B be the area of the circle circumscribed about the triangle. Find A/B

 

Let the perimeter be 12p

So the sides of the square are 3p each and the sides of the triangle are 4p each

 

\(A \div B = \frac{4.5\pi p^2}{1}\div \frac{16p^2\pi}{3}\\ A \div B = \frac{4.5\pi p^2}{1}\times \frac{3}{16p^2\pi}\\ \frac{A}{B} = \frac{4.5}{1}\times \frac{3}{16}\\ \frac{A}{B} = \frac{4.5*3}{16}\\ \frac{A}{B} = \frac{13.5}{16}\\ \frac{A}{B} = \frac{27}{32}\\ \)

Melody  Nov 21, 2015
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2+0 Answers

 #1
avatar+91038 
+15
Best Answer

A square and an equilateral triangle have the same perimeter. Let Abe the area of the circle circumscribed about the square and B be the area of the circle circumscribed about the triangle. Find A/B

 

Let the perimeter be 12p

So the sides of the square are 3p each and the sides of the triangle are 4p each

 

\(A \div B = \frac{4.5\pi p^2}{1}\div \frac{16p^2\pi}{3}\\ A \div B = \frac{4.5\pi p^2}{1}\times \frac{3}{16p^2\pi}\\ \frac{A}{B} = \frac{4.5}{1}\times \frac{3}{16}\\ \frac{A}{B} = \frac{4.5*3}{16}\\ \frac{A}{B} = \frac{13.5}{16}\\ \frac{A}{B} = \frac{27}{32}\\ \)

Melody  Nov 21, 2015
 #2
avatar+78719 
+10

Very nice, Melody!!!!........great diagrams, too....!!!!!

 

Here's an alternative method.....

 

Let the side of the square = (1/4)P.....then the diagonal = [ sqrt(2)/4] P    ....and the radius = [sqrt(2)/8] P

 

Let the side of the equilateral triangle = (1/3)P.....then by the Law of Sines, the radius of  circle  B can be forund thusly :

 

(1/3)P / sin(120) = r / sin (30)

 

r =  [(1/2) * (1/3)P] / [sqrt(3)/2] =  [(1/3)P] / sqrt(3)

 

And, by Euclid, circles are to each other as the squares of their radiuses....so we have....

 

A / B   =  ( [sqrt(2)/8] P )^2  /  (   [(1/3)P] / sqrt(3) )^2    =   [ 1/32 ]  /  [ 1/27]  =  27/32

 

 

cool cool cool

CPhill  Nov 21, 2015
edited by CPhill  Nov 21, 2015

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