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Find the circumradius of triangle JKL if KL=16 and JK=JL=17

Mellie  Nov 20, 2015

Best Answer 

 #1
avatar+78739 
+15

From the associated diagram below, it is obvious that the radius of the circle on which each of the vertices of triangle JKL lie will be located at (8, n)........where n is the y coordinate that we're looking for....let's call this point "O"

 

 

We can equate distances here to solve for n

 

The distance from K to O    =  sqrt [ 8^2 + n^2]   = sqrt [ 64 +n^2 ]

 

And the distance from  J to O   = sqrt  [ (15 - n)^2]  =  15 - n

 

So .......sqrt [ 64] + n^2] = 15 - n     ...... square both sides

 

64 + n^2  =  225 - 30n + n^2

 

30n  = 225 - 64

 

30n =  161

 

n = 161 / 30

 

So, "O"  =  (8, 161/30)

 

And the distance from J  to O   =  15 - 161/30   =  [450 -  161] / 30   =  289/30 .....  this is the circumradius......!!!

 

And the equation of the circle through all three points is :

 

(x - 8)^2 + (y - 161/30)^2   = (289/  30)^2

 

P.S.  = There is a "formula" for finding the circumradius of a triangle  =    [ a* b * c ] / [ 4 A]    where a,b,c are the sides of the triangle and A is the area......

 

Using this, we have   [ 16 * 17^2]  / [ 4 * 120]  = 289 / 30   .....the same result as mine  !!!!

 

Here's the proof of this "formula"...it's surprisingly simple !!!!........http://www.artofproblemsolving.com/wiki/index.php/Circumradius

 

 

cool cool cool

CPhill  Nov 21, 2015
edited by CPhill  Nov 21, 2015
edited by CPhill  Nov 21, 2015
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1+0 Answers

 #1
avatar+78739 
+15
Best Answer

From the associated diagram below, it is obvious that the radius of the circle on which each of the vertices of triangle JKL lie will be located at (8, n)........where n is the y coordinate that we're looking for....let's call this point "O"

 

 

We can equate distances here to solve for n

 

The distance from K to O    =  sqrt [ 8^2 + n^2]   = sqrt [ 64 +n^2 ]

 

And the distance from  J to O   = sqrt  [ (15 - n)^2]  =  15 - n

 

So .......sqrt [ 64] + n^2] = 15 - n     ...... square both sides

 

64 + n^2  =  225 - 30n + n^2

 

30n  = 225 - 64

 

30n =  161

 

n = 161 / 30

 

So, "O"  =  (8, 161/30)

 

And the distance from J  to O   =  15 - 161/30   =  [450 -  161] / 30   =  289/30 .....  this is the circumradius......!!!

 

And the equation of the circle through all three points is :

 

(x - 8)^2 + (y - 161/30)^2   = (289/  30)^2

 

P.S.  = There is a "formula" for finding the circumradius of a triangle  =    [ a* b * c ] / [ 4 A]    where a,b,c are the sides of the triangle and A is the area......

 

Using this, we have   [ 16 * 17^2]  / [ 4 * 120]  = 289 / 30   .....the same result as mine  !!!!

 

Here's the proof of this "formula"...it's surprisingly simple !!!!........http://www.artofproblemsolving.com/wiki/index.php/Circumradius

 

 

cool cool cool

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

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