# A triangle has corners at #(1 , 2 )#, #(5 ,7 )#, and #(9 ,5 )#. What is the radius of the triangle's inscribed circle?

Radius of inscribed circle

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To find the radius ( r ) of the inscribed circle of a triangle given its vertices ( (x_1, y_1) ), ( (x_2, y_2) ), and ( (x_3, y_3) ), you can use the formula:

[ r = \frac{2 \times \text{Area}}{ \text{Perimeter}} ]

where

[ \text{Area} = \sqrt{s \times (s-a) \times (s-b) \times (s-c)} ]

[ s = \frac{a + b + c}{2} ]

[ \text{Perimeter} = a + b + c ]

[ a = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ]

[ b = \sqrt{(x_3 - x_2)^2 + (y_3 - y_2)^2} ]

[ c = \sqrt{(x_3 - x_1)^2 + (y_3 - y_1)^2} ]

Using the given points ( (1, 2) ), ( (5, 7) ), and ( (9, 5) ):

[ a = \sqrt{(5 - 1)^2 + (7 - 2)^2} = \sqrt{16 + 25} = \sqrt{41} ]

[ b = \sqrt{(9 - 5)^2 + (5 - 7)^2} = \sqrt{16 + 4} = \sqrt{20} ]

[ c = \sqrt{(9 - 1)^2 + (5 - 2)^2} = \sqrt{64 + 9} = \sqrt{73} ]

[ s = \frac{\sqrt{41} + \sqrt{20} + \sqrt{73}}{2} ]

[ \text{Area} = \sqrt{s \times (s - \sqrt{41}) \times (s - \sqrt{20}) \times (s - \sqrt{73})} ]

[ \text{Perimeter} = \sqrt{41} + \sqrt{20} + \sqrt{73} ]

[ r = \frac{2 \times \text{Area}}{\text{Perimeter}} ]

[ r = \frac{2 \times \sqrt{s \times (s - \sqrt{41}) \times (s - \sqrt{20}) \times (s - \sqrt{73})}}{\sqrt{41} + \sqrt{20} + \sqrt{73}} ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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