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

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

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Answer 1 · 2017-02-24T18:59:05

$\frac{\sqrt{5}}{2 + \sqrt{2}}$ $sqrt5/(2+sqrt2)$

Explanation:

If A is (5,2) B is (2,3) and C is (3,1), then

Side $A B = \sqrt{{(5 - 2)}^{2} + {(2 - 3)}^{2}} = \sqrt{10}$ $AB= sqrt((5-2)^2 +(2-3)^2)=sqrt10$

Side $B C = \sqrt{{(2 - 3)}^{2} + {(3 - 1)}^{2}} = \sqrt{5}$ $BC= sqrt ((2-3)^2 +(3-1)^2)=sqrt5$

Side $A C = \sqrt{{(5 - 3)}^{2} + {(2 - 1)}^{2}} = \sqrt{5}$ $AC= sqrt((5-3)^2 +(2-1)^2)= sqrt5$

Since $A C^{2} + B C^{2} = A B^{2}$ $AC^2+BC^2 =AB^2$ , the triangle is a right triangle. Its Area would be $\frac{1}{2} \sqrt{5} \cdot \sqrt{5} = \frac{5}{2}$ $1/2 sqrt5 *sqrt5= 5/2$

Sum of the $s i d e s = \sqrt{5} + \sqrt{5} + \sqrt{10} = \sqrt{5} (2 + \sqrt{2})$ $sides =sqrt5 +sqrt5 +sqrt10=sqrt5(2+sqrt2)$

radius of incircle would be $2 \frac{A r e a o f △}{\sum s i d e s o f △}$ $2 (Area of triangle) / (sum sides oftriangle)$

= $2 \cdot \frac{5}{2} \div \sqrt{5} (2 + \sqrt{2}) = \frac{\sqrt{5}}{2 + \sqrt{2}}$ $2* 5/2 div sqrt5 (2+sqrt2)= sqrt5/(2+sqrt2)$

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

1 Answer

Explanation:

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

1 Answer

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