A triangle has corners at #(7 , 9 )#, #(3 ,7 )#, and #(4 ,8 )#. What is the radius of the triangle's inscribed circle?

Question

A triangle has corners at #(7  ,  9    )#, #(3   ,7  )#, and #(4  ,8   )#.  What is the radius of the triangle's inscribed circle?

Autumn Armstrong · Accepted Answer

Incenter radius #r = A_t / s = 1 / 4.5243 = color(green)(0.221)#

Given the coordinates of the three vertices of a triangle ABC,
the coordinates of the incenter O are

Given A(7,9), B(3,7), C(4,8)

Using distance formula we can calculate a, b, c.

#a = sqrt((4-3)^2 + (8-7)^2) = 1.4142#

#b = sqrt((9-8)^2 + (7-4)^2) = 3.1623#

#c = sqrt((7-9)^2 + (3-7)^2) = 4.4721#

Semi perimeter #s = (a + b + c)/2 = (1.4142 + 3.1623 + 4.4721)/2 = 4.5243#

Area of triangle #A_t = sqrt(s (s-a) (s-b) (s-c))#

#A_t = sqrt(4.5243(4.5243-1.4142)(4.5243-3.1623)(4.5243-4.4721)) = 1#

Incenter radius #r = A_t / s = 1 / 4.5243 = color(green)(0.221)#

Claire Akin · Answer

undefined To find the radius of the inscribed circle in a triangle, use the formula:

$$ r = \frac{A}{s} $$

Where $ A $ is the area of the triangle and $ s $ is the semi-perimeter of the triangle.

$$ A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| $$

$$ s = \frac{a + b + c}{2} $$

Given the coordinates:

$ (x_1, y_1) = (7, 9) $

$ (x_2, y_2) = (3, 7) $

$ (x_3, y_3) = (4, 8) $

$$ a = \sqrt{(x_2 - x_3)^2 + (y_2 - y_3)^2} $$
$$ b = \sqrt{(x_3 - x_1)^2 + (y_3 - y_1)^2} $$
$$ c = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} $$

Substitute the coordinates into the formulas and calculate:

$$ A = \frac{1}{2} |7(7 - 8) + 3(8 - 9) + 4(9 - 7)| = \frac{1}{2} |7 - 3 + 8| = \frac{1}{2} \times 12 = 6 $$

$$ a = \sqrt{(3 - 4)^2 + (7 - 8)^2} = \sqrt{1 + 1} = \sqrt{2} $$
$$ b = \sqrt{(4 - 7)^2 + (8 - 9)^2} = \sqrt{9 + 1} = \sqrt{10} $$
$$ c = \sqrt{(7 - 3)^2 + (9 - 7)^2} = \sqrt{16 + 4} = \sqrt{20} $$

$$ s = \frac{\sqrt{2} + \sqrt{10} + \sqrt{20}}{2} $$

Now, plug the values of $ A $ and $ s $ into the formula for the radius of the inscribed circle:

$$ r = \frac{6}{\frac{\sqrt{2} + \sqrt{10} + \sqrt{20}}{2}} $$

$$ r = \frac{12}{\sqrt{2} + \sqrt{10} + \sqrt{20}} $$

$$ r ≈ \frac{12}{1.41 + 3.16 + 4.47} $$

$$ r ≈ \frac{12}{9.04} $$

$$ r ≈ 1.33 $$

So, the radius of the triangle's inscribed circle is approximately $ 1.33 $ units.