A triangle has corners at #(3 , 4 )#, #(6 ,3 )#, and #(5 ,8 )#. What is the radius of the triangle's inscribed circle?

Answer 1

#R=sqrt{2* (sqrt5 - sqrt13 + 3) * (sqrt5 + sqrt13 - 3) * (-sqrt5 + sqrt13 + 3)}/{sqrt{sqrt5 + sqrt13 + 3}}#

First, we find the length of each side of the triangle using the Pythagorean Theorem.

#sqrt{(6 - 3)^2 + (3 - 4)^2} = sqrt10#
#sqrt{(6 - 5)^2 + (3 - 8)^2} = sqrt26#
#sqrt{(3 - 5)^2 + (4 - 8)^2} = 3sqrt2#

There is a short way to find the radius of the incircle. It is given by the following formula.

#"Radius of Incircle" = frac{2 * "Area of "Delta}{"Perimeter of "Delta}#
The perimeter is #sqrt10 + sqrt26 + 3sqrt2#.

The area is found quickly using Heron's Formula.

The semi-perimeter is #frac{sqrt5 + sqrt13 + 3}{sqrt2}#

The area is

#sqrt{(frac{sqrt5 + sqrt13 + 3}{sqrt2}) * (frac{sqrt5 - sqrt13 + 3}{sqrt2}) * (frac{sqrt5 + sqrt13 - 3}{sqrt2}) * (frac{-sqrt5 + sqrt13 + 3}{sqrt2})}#
#=sqrt{(sqrt5 + sqrt13 + 3) * (sqrt5 - sqrt13 + 3) * (sqrt5 + sqrt13 - 3) * (-sqrt5 + sqrt13 + 3)}/2#

The radius is

#=sqrt{2* (sqrt5 - sqrt13 + 3) * (sqrt5 + sqrt13 - 3) * (-sqrt5 + sqrt13 + 3)}/{sqrt{sqrt5 + sqrt13 + 3}}#
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Answer 2

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

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

First, calculate the lengths of the sides of the triangle using the distance formula:

[ AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} ] [ BC = \sqrt{(x_3 - x_2)^2 + (y_3 - y_2)^2} ] [ AC = \sqrt{(x_3 - x_1)^2 + (y_3 - y_1)^2} ]

Then, compute the semi-perimeter of the triangle:

[ s = \frac{AB + BC + AC}{2} ]

Next, find the area of the triangle using Heron's formula:

[ \text{Area} = \sqrt{s \times (s - AB) \times (s - BC) \times (s - AC)} ]

Now, you have both the area and the perimeter of the triangle. Plug these values into the formula for the radius of the inscribed circle to get ( r ).

Given the coordinates of the triangle's vertices:

( A(3, 4) ), ( B(6, 3) ), and ( C(5, 8) )

Using the distance formula, calculate the lengths of the sides:

( AB = \sqrt{(6 - 3)^2 + (3 - 4)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} )

( BC = \sqrt{(5 - 6)^2 + (8 - 3)^2} = \sqrt{(-1)^2 + 5^2} = \sqrt{1 + 25} = \sqrt{26} )

( AC = \sqrt{(5 - 3)^2 + (8 - 4)^2} = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} )

Calculate the semi-perimeter:

( s = \frac{AB + BC + AC}{2} = \frac{\sqrt{10} + \sqrt{26} + 2\sqrt{5}}{2} )

Compute the area of the triangle using Heron's formula:

( \text{Area} = \sqrt{s \times (s - AB) \times (s - BC) \times (s - AC)} )

Now that you have the area and the perimeter, use the formula for the radius of the inscribed circle:

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

Plug in the values to find ( r ).

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Answer from HIX Tutor

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.

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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