The bases of the truncated pyramid are rectangles, and the points of intersection of the base diagonals

The bases of the truncated pyramid are rectangles, and the points of intersection of the base diagonals are located on the same perpendicular to the base plane. The sides of one rectangle are 54 cm and 30 cm; the perimeter of another rectangle is 112 cm; the distance between their planes is 12 cm. Determine the lateral surface of this truncated pyramid.

Since the intersection points of the diagonals of the bases lie on one perpendicular to their planes, the pyramid is symmetrical, its sides are isosceles trapeziums, and the rectangles at the base are similar.

Let A1D1 = X cm, and A1B1 = Y cm.

Then:

2 * X + 2 * Y = 112 cm.

AD / A1D1 = AB / A1В1.

54 / X = 30 / Y.

30 * X = 54 * Y.

Let’s solve a system of two equations.

X = (112 – 2 * Y) / 2 = 56 – Y.

30 * (56 – Y) = 54 * Y.

1680 – 30 * Y = 54 * Y.

Y = 1680/84 = 20 cm.

X = 56 – 20 = 36 cm.

A1D1 = 36 cm.

A1B1 = 20 cm.

Consider a right-angled triangle Н1НМ, and by the Pythagorean theorem define the hypotenuse НМ.

НМ2 = НН12 + Н1М2 = 122 + (27 – 18) 2 = 144 + 81 = 225.

НM = 15 cm.

Consider a right-angled triangle K1KP, and by the Pythagorean theorem we define the hypotenuse KP.

KP^2 = KK1^2 + K1P^2 = 12^2 + (15 – 10)^2 = 144 + 25 = 169.

KР = 13 cm.

The area of ​​the side area will be equal to the sum of the areas of the trapezoids that make up the sides.

S side = 2 * (BP + A1D1) * KР / 2 + 2 * (СD + C1D1) * НM / 2 = 2 * (54 + 36) * 13/2 + 2 * (30 + 20) * 15/2 = 1170 + 750 = 1920 cm2.

Answer: The lateral surface area is 1920 cm2.