A triangle has corners at $(2, 6)$ $(2 , 6 )$ , $(8, 2)$ $(8 ,2 )$ , and $(1, 3)$ $(1 ,3 )$ . What is the radius of the triangle's inscribed circle?

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A triangle has corners at $(2, 6)$ $(2 , 6 )$ , $(8, 2)$ $(8 ,2 )$ , and $(1, 3)$ $(1 ,3 )$ . What is the radius of the triangle's inscribed circle?

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Answer 1 · 2018-06-21T05:36:25

$Radius of incircle r = \frac{A_{t}}{s} = \frac{10.99}{8.72} = 1.26 units$ $color(orange)("Radius of incircle " r = A_t / s = 10.99 / 8.72 = 1.26 " units"$

Explanation:

![http://mathibayon.blogspot.com/2015/01/http://derivation-of-formula-for-radius-of-incircle.html](https://useruploads.socratic.org/qByQYJn5SEeAPjtKhB4j_incircle%20radius.png)

$Incircle radius r = \frac{A_{t}}{s}$ $"Incircle radius " r = A_t / s$

$A (2, 6), B (8, 2), C (1, 3)$ $A(2,6), B(8,2), C(1,3)$

$a = \sqrt{{(8 - 1)}^{2} + {(2 - 3)}^{2}} = 7.07$ $a = sqrt((8-1)^2 + (2-3)^2) = 7.07$

$b = \sqrt{{(1 - 2)}^{2} + {(3 - 6)}^{2}} = 3.16$ $b = sqrt((1-2)^2 + (3-6)^2) = 3.16$

$c = \sqrt{{(2 - 8)}^{2} + {(6 - 2)}^{2}} = 7.21$ $c = sqrt((2-8)^2 + (6-2)^2) = 7.21$

$Semi-perimeter s = \frac{a + b + c}{2} = \frac{7.07 + 3.16 + 7.21}{2} = 8.72$ $"Semi-perimeter " s = (a + b + c) / 2 = (7.07 + 3.16 + 7.21) / 2 = 8.72$

$A_{t} = \sqrt{s (s - a) s - b} (s - c))$ $"A_t = sqrt(s (s-a) s-b) (s-c))$

$A_{t} = \sqrt{8.72 (8.72 - 7.07) (8.72 - 3.16) (8.72 - 7.21)} = 10.99$ $A_t = sqrt(8.72 (8.72-7.07) (8.72 - 3.16) (8.72 - 7.21)) = 10.99$

$Radius of incircle r = \frac{A_{t}}{s} = \frac{10.99}{8.72} = 1.26 units$ $color(orange)("Radius of incircle " r = A_t / s = 10.99 / 8.72 = 1.26 " units"$

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A triangle has corners at $(2, 6)$ $(2 , 6 )$ , $(8, 2)$ $(8 ,2 )$ , and $(1, 3)$ $(1 ,3 )$ . What is the radius of the triangle's inscribed circle?

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