# The sides of the bases of a regular truncated quadrangular pyramid are 4m and 20m

**The sides of the bases of a regular truncated quadrangular pyramid are 4m and 20m, and the length of the side rib is 10m. Find the total surface area of the prism.**

Let us determine the lengths of the diagonals of the squares that lie at the bases of the truncated pyramid.

AC = AD * √2 = 20 * √2 cm.

A1C1 = A1D1 * √2 = 4 * √2 cm.

The diagonals of the bases form an isosceles pyramid АА1С1С.

Let’s draw the height А1К, which on the diagonal АС cuts off the segment АС, the length of which is equal to the half-difference АС and А1С1.

AK = (AC – A1C1) / 2 = (20 * √2 – 4 * √2) / 2 = 8 * √2 cm.

Let’s draw the apothem of the PH and the height of the RM, which is equal to A1K = 8 * √2.

Segment OH = AD / 2 = 20/2 = 10 cm.

Segment ОМ = А1D1 / 2 = 4/2 = 2 cm.

Then the segment MH = OH – OM = 10 – 2 = 8 cm.

From the right-angled triangle РМН, according to the Pythagorean theorem, we define the hypotenuse РН.

PH ^ 2 = PM ^ 2 + MH ^ 2 = (8 * √2) ^ 2 + 8 ^ 2 = 128 + 64 = 192.

PH = 8 * √3 cm.

Let us determine the area of the lateral surface.

Side = 4 * (CD + C1D1) * PH / 2 = 4 * (20 + 4) * 8 * √3 / 2 = 384 * √3 cm2.

Sbn = AD ^ 2 + A1D1 ^ 2 = 400 + 16 = 416 cm2.

S floor = S side + S main = 384 * √3 + 416 = 32 * (12 * √3 + 13) cm2.

Answer: The total area is 32 * (12 * √3 + 13) cm2.