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

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

Nov 20, 2015

#1
+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   Nov 21, 2015
edited by CPhill  Nov 21, 2015
edited by CPhill  Nov 21, 2015

#1
+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   CPhill Nov 21, 2015
edited by CPhill  Nov 21, 2015
edited by CPhill  Nov 21, 2015