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

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

Mellie  Nov 20, 2015

#1
+81014
+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

CPhill  Nov 21, 2015
Sort:

#1
+81014
+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

CPhill  Nov 21, 2015
#2
+91451
+5