A triangle has corners at $(4, 5)$ $(4 , 5 )$ , $(3, 2)$ $(3 ,2 )$ , and $(1, 3)$ $(1 ,3 )$ . What is the radius of the triangle's inscribed circle?

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A triangle has corners at $(4, 5)$ $(4 , 5 )$ , $(3, 2)$ $(3 ,2 )$ , and $(1, 3)$ $(1 ,3 )$ . What is the radius of the triangle's inscribed circle?

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Answer 1 · 2018-06-22T09:41:24

$Radius of inscribed circle r = \frac{A_{t}}{s} = \frac{3.62}{4.529} = 0.8 units$ $color(cyan)("Radius of inscribed circle " r = A_t / s = 3.62/ 4.529 = 0.8 " units"$

Explanation:

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

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

$A (4, 5), B (3, 2), C (1, 3)$ $A(4,5), B(3,2), C(1,3)$

$a = \sqrt{{(3 - 1)}^{2} + {(2 - 3)}^{2}} = 2.236$ $a = sqrt((3-1)^2 + (2-3)^2) = 2.236$

$b = \sqrt{{(1 - 4)}^{2} + {(3 - 5)}^{2}} = 3.606$ $b = sqrt((1-4)^2 + (3-5)^2) = 3.606$

$c = \sqrt{{(4 - 3)}^{2} + {(5 - 2)}^{2}} = 3.162$ $c = sqrt((4-3)^2 + (5-2)^2) = 3.162$

$Semi-perimeter s = \frac{a + b + c}{2} = \frac{2.236 + 3.606 + 3.162}{2} = 4, 529$ $"Semi-perimeter " s = (a + b + c) / 2 = (2.236 + 3.606 + 3.162) / 2 = 4,529$

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

$A_{t} = \sqrt{4.529 (4.529 - 2.236) (4.529 - 3.606) (4.529 - 3.162)} = 3.62$ $A_t = sqrt(4.529 (4.529-2.236) (4.529 - 3.606) (4.529 - 3.162)) = 3.62$

$Radius of inscribed circle r = \frac{A_{t}}{s} = \frac{3.62}{4.529} = 0.8 units$ $color(cyan)("Radius of inscribed circle " r = A_t / s = 3.62/ 4.529 = 0.8 " units"$

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A triangle has corners at $(4, 5)$ $(4 , 5 )$ , $(3, 2)$ $(3 ,2 )$ , and $(1, 3)$ $(1 ,3 )$ . What is the radius of the triangle's inscribed circle?

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Explanation:

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Explanation:

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