A triangle has corners at $(2, 3)$ $(2 , 3 )$ , $(1, 5)$ $(1 ,5 )$ , and $(6, 7)$ $(6 ,7 )$ . What is the radius of the triangle's inscribed circle?

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A triangle has corners at $(2, 3)$ $(2 , 3 )$ , $(1, 5)$ $(1 ,5 )$ , and $(6, 7)$ $(6 ,7 )$ . What is the radius of the triangle's inscribed circle?

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Answer 1 · 2018-07-29T12:27:15

$Radius of in-circle = r = \frac{A_{t}}{s} \approx 0.9038$ $color(orange)("Radius of in-circle "= r = A_t / s ~~ 0.9038$

Explanation:

$A (2, 3), B (1, 5), C (6, 7)$ $A(2, 3), B(1, 5), C(6, 7)$

$c = \sqrt{{(2 - 1)}^{2} + {(3 - 5)}^{2}} \approx \sqrt{5}$ $c = sqrt((2-1)^2 + (3-5)^2) ~~ sqrt 5$

$a = \sqrt{{(1 - 6)}^{2} + {(5 - 7)}^{2}} \approx \sqrt{29}$ $a= sqrt ((1-6)^2 + (5-7)^2) ~~ sqrt 29$

$b = \sqrt{{(6 - 2)}^{2} + {(7 - 3)}^{2}} \approx \sqrt{32}$ $b = sqrt((6-2)^2 + (7-3)^2) ~~ sqrt 32$

Semi perimeter $s = \frac{a + b + c}{2}$ $s = (a + b + c)/2$

$s = \frac{\sqrt{5} + \sqrt{29} + \sqrt{32}}{2} = 6.639$ $s = (sqrt 5 + sqrt 29 + sqrt 32 ) / 2 = 6.639$

Area of triangle $A_{t} = \sqrt{s (s - a) (s - b) (s - c)}, using Heron's formula$ $A_t = sqrt(s (s-a) (s-b) (s-c)), " using Heron's formula"$

$A_{t} = \sqrt{6.639 (6.639 - \sqrt{5}) (6.639 - \sqrt{29}) (6.639 - \sqrt{32})} \approx 6$ $A_t = sqrt(6.639 (6.639- sqrt 5) (6.639-sqrt 29) (6.639-sqrt 32)) ~~ 6$

$Radius of in-circle = r = \frac{A_{t}}{s} = \frac{6}{6.639} \approx 0.9038$ $color(orange)("Radius of in-circle "= r = A_t / s = 6 / 6.639 ~~ 0.9038$

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A triangle has corners at $(2, 3)$ $(2 , 3 )$ , $(1, 5)$ $(1 ,5 )$ , and $(6, 7)$ $(6 ,7 )$ . What is the radius of the triangle's inscribed circle?

1 Answer

Explanation:

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A triangle has corners at (2,3)(2 , 3 ), (1,5)(1 ,5 ), and (6,7)(6 ,7 ). What is the radius of the triangle's inscribed circle?

1 Answer

Explanation:

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A triangle has corners at $(2, 3)$ $(2 , 3 )$ , $(1, 5)$ $(1 ,5 )$ , and $(6, 7)$ $(6 ,7 )$ . What is the radius of the triangle's inscribed circle?