A 9cm high box is in the form of a frustum of a regular square pyramid.what is the valume of the box if the area based are 4cm^2 and 36cm^2? ?

To find the volume of the frustum of a square pyramid, we can use the formula:

V = (1/3)h(A1 + A2 + √(A1 * A2))

Where:
V = volume of the frustum
h = height of the frustum (9 cm)
A1 = area of the lower base (4 cm^2)
A2 = area of the upper base (36 cm^2)

Plugging in the values:

V = (1/3)(9)(4 + 36 + √(4 * 36))
V = (1/3)(9)(40 + √144)
V = (1/3)(9)(40 + 12)
V = (1/3)(9)(52)
V = 156 cm^3

Therefore, the volume of the box is 156 cm^3.

Let the side of the lower base is 2a cm and the side of the upper base is 2b cm.

Then Volume of the frustum = (1/3) - (4a^2 + 4ab + 4b^2) - h

Given that, h = 9cm
Area of the lower base = 4a^2 = 4cm^2
Area of the upper base = 4b^2 = 36cm^2
4a^2 = 4cm^2 => a = 1cm
4b^2 = 36cm^2 => b = 3cm

Therefore, Volume of the frustum = (1/3) - (4 - 1^2 + 4 - 1 - 3 + 4 - 3^2) - 9
= (1/3) - (4 + 12 + 36) - 9
= (1/3) - 52 - 9
= 52 - 3
= 156 cm^3

Hence, the volume of the box is 156 cm^3.