A triangle has corners at $(6, 3)$ $(6 , 3 )$ , $(1, 5)$ $(1 ,5 )$ , and $(2, 5)$ $(2 ,5 )$ . What is the radius of the triangle's inscribed circle?

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A triangle has corners at $(6, 3)$ $(6 , 3 )$ , $(1, 5)$ $(1 ,5 )$ , and $(2, 5)$ $(2 ,5 )$ . What is the radius of the triangle's inscribed circle?

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Answer 1 · 2018-06-22T08:11:38

$Radius of incircle r = \frac{A_{t}}{s} = \frac{1.55}{5.92} \approx 0.26 units$ $color(indigo)("Radius of incircle " r = A_t / s = 1.55 / 5.92 ~~ 0.26 " units"$

Explanation:

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

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

$A (6, 2), B (1, 5), C (2, 5)$ $A(6,2), B(1,5), C(2,5)$

$a = \sqrt{{(1 - 2)}^{2} + {(5 - 5)}^{2}} = 1$ $a = sqrt((1-2)^2 + (5-5)^2) = 1$

$b = \sqrt{{(2 - 6)}^{2} + {(5 - 2)}^{2}} = 5$ $b = sqrt((2-6)^2 + (5-2)^2) = 5$

$c = \sqrt{{(6 - 1)}^{2} + {(2 - 5)}^{2}} = 5.83$ $c = sqrt((6-1)^2 + (2-5)^2) = 5.83$

$Semi-perimeter s = \frac{a + b + c}{2} = \frac{1 + 5 + 5.83}{2} = 5.92$ $"Semi-perimeter " s = (a + b + c) / 2 = (1 + 5 + 5.83) / 2 = 5.92$

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

$A_{t} = \sqrt{5.92 (5.92 - 1) (5.92 - 5) (5.92 - 5.83)} = 1.55$ $A_t = sqrt(5.92 (5.92-1) (5.92 - 5) (5.92 - 5.83)) = 1.55$

$Radius of incircle r = \frac{A_{t}}{s} = \frac{1.55}{5.92} \approx 0.26 units$ $color(indigo)("Radius of incircle " r = A_t / s = 1.55 / 5.92 ~~ 0.26 " units"$

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A triangle has corners at $(6, 3)$ $(6 , 3 )$ , $(1, 5)$ $(1 ,5 )$ , and $(2, 5)$ $(2 ,5 )$ . What is the radius of the triangle's inscribed circle?

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A triangle has corners at (6,3)(6 , 3 ), (1,5)(1 ,5 ), and (2,5)(2 ,5 ). What is the radius of the triangle's inscribed circle?

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A triangle has corners at $(6, 3)$ $(6 , 3 )$ , $(1, 5)$ $(1 ,5 )$ , and $(2, 5)$ $(2 ,5 )$ . What is the radius of the triangle's inscribed circle?