The base of a triangular pyramid is a triangle with corners at #(4 ,2 )#, #(3 ,6 )#, and #(7 ,5 )#. If the pyramid has a height of #6 #, what is the pyramid's volume?

To find the volume of the triangular pyramid, we'll first calculate the area of its base triangle using the formula for the area of a triangle. Then, we'll use the formula for the volume of a pyramid.

Given the coordinates of the base triangle corners: (4, 2), (3, 6), and (7, 5), we can use these to find the base triangle's area.

Using the coordinates, the lengths of the base triangle's sides can be calculated:

Side 1: (\sqrt{(4-3)^2 + (2-6)^2} = \sqrt{1^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17})

Side 2: (\sqrt{(3-7)^2 + (6-5)^2} = \sqrt{(-4)^2 + 1^2} = \sqrt{16 + 1} = \sqrt{17})

Side 3: (\sqrt{(7-4)^2 + (5-2)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2})

Using Heron's formula, the area ((A_b)) of the base triangle can be calculated:

(s = \frac{(\sqrt{17} + \sqrt{17} + 3\sqrt{2})}{2} = \frac{2\sqrt{17} + 3\sqrt{2}}{2})

(A_b = \sqrt{s(s - \sqrt{17})(s - \sqrt{17})(s - 3\sqrt{2})})

Now, let's find the volume of the pyramid using the formula:

(V = \frac{1}{3} \times A_b \times h)

Substitute the values:

(V = \frac{1}{3} \times \sqrt{s(s - \sqrt{17})(s - \sqrt{17})(s - 3\sqrt{2})} \times h)

Given that the height (h = 6), we can now compute the volume of the pyramid.