Geometry help?

Question

Find the volume of a frustum of a right pyramid whose lower base is a square with a side 5in, whose upper base is a square with a side 3in, and whose altitude is 12in.

Answer 1

I assume the altitude given is the altitude of the entire pyramid?

First, we need the altitude of the point that is cut off...

Making a right triangle from the altitude to one side of the bottom base, we have the proportion:

12 / 2.5 = x / 1.5
x = 7.2 (That's the altitude of the point that is cut off.)

So...

The Volume of the whole pyramid is:
V = (1/3)(5*5)(12) = 100

The Volume of the top part that is cut off is:
V = (1/3)(3*3)(7.2) = 21.6

So the volume of the frustum is:
100 - 21.6 = 78.4

Answer 2

V=1/6*h[a^2+(a+a_1)^2+a_1^2] where a and a_1 are the sides of the lover and upper base
V=2(25+64+9)= 196 cubic inches

Answer 3

78.4
Please give me best answer thanks!