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

Mellie  Nov 20, 2015

Best Answer 

 #1
avatar+90023 
+15

Here's the diagram.......

 

 

The inradius  will be located at (8,n)   where n is the y coordinate we are looking for....let's call this "O"

 

The equation of the line containing segment KL  is  y = 0

And the equation of the line that contains segment JK   is  y = (15/8)x →  15x - 8y = 0

 

And  "O" =  (8,n) will be equidistant from JK and KL

 

Using the "formula"    for the distance from a point to a line, we have

 

abs(15(8) - 8n)  / sqrt [ 15^2 + (-8)^2 ]      =  abs ( n )     

 

And we will assume that both n  and 15(8) - 8n > 0....so we have

 

[120 - 8n] / 17   = n     

 

[120 - 8n ] = 17n

 

120  = 25n

 

n = 120/25 =  24/5

 

And the distance ffrom this point to the line y = 0   is just 24/5........and this is the inradius

 

The equation of the incircle of triangle JKL  is   (x - 8)^2 + (y - 24/5)^2  = (24/5)^2

 

 

 

cool cool cool

CPhill  Nov 21, 2015
 #1
avatar+90023 
+15
Best Answer

Here's the diagram.......

 

 

The inradius  will be located at (8,n)   where n is the y coordinate we are looking for....let's call this "O"

 

The equation of the line containing segment KL  is  y = 0

And the equation of the line that contains segment JK   is  y = (15/8)x →  15x - 8y = 0

 

And  "O" =  (8,n) will be equidistant from JK and KL

 

Using the "formula"    for the distance from a point to a line, we have

 

abs(15(8) - 8n)  / sqrt [ 15^2 + (-8)^2 ]      =  abs ( n )     

 

And we will assume that both n  and 15(8) - 8n > 0....so we have

 

[120 - 8n] / 17   = n     

 

[120 - 8n ] = 17n

 

120  = 25n

 

n = 120/25 =  24/5

 

And the distance ffrom this point to the line y = 0   is just 24/5........and this is the inradius

 

The equation of the incircle of triangle JKL  is   (x - 8)^2 + (y - 24/5)^2  = (24/5)^2

 

 

 

cool cool cool

CPhill  Nov 21, 2015
 #2
avatar+93683 
+5

Nice answer Chris  laugh 

Melody  Nov 22, 2015

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