ABCDA1B1C1D1 is a regular truncated pyramid. The side of the upper base is 2 cm

| January 31, 2021 | education

ABCDA1B1C1D1 is a regular truncated pyramid. The side of the upper base is 2 cm, the lower base is 6 cm, and the side is 7 cm. Find the surface area of the pyramid (squares at the bases).

Let’s calculate the areas of the bases of the truncated pyramid.

Sop1 = AB ^ 2 = 6 ^ 2 = 36 cm2.

Sbn2 = A1B12 = 4 cm2.

Let us determine the lengths of the diagonals A1C1 and AC.

А1С1 = А1В1 * √2 = 2 * √2 cm.

AC = AB * √2 = 6 * √2 cm.

Quadrangle MA1O1O is a rectangle, then MO = A1O1 = A1C1 / 2 = 2 * √2 / 2 = √2 cm, A1M = OO1 = 7 cm.

Then AM = AO – OM = (AC / 2) – OM = 3 * √2 – √2 = 2 * √2 cm.

In a right-angled triangle АА1М, according to the Pythagorean theorem, we determine the length of the hypotenuse АА1.

AA1 ^ 2 = A1M ^ 2 + AM ^ 2 = 49 + 8 = 57.

Let us stand the height A1H to the AB side.

The side faces of the pyramid are isosceles trapezoid, then AH = (AB – A1B1) / 2 = 2 cm.

In a right-angled triangle АА1Н, according to the Pythagorean theorem, А1Н ^ 2 = АА1 ^ 2 – АН ^ 2 = 57 – 4 = 53.

A1H = √53 cm.

Then Saa1b1v = (A1B1 + AB) * A1H / 2 = 8 * √53 / 2 = 4 * √53 cm2.

Then Sside = 4 * Saa1v1v = 16 * √53 cm2.

Spov = S side + Sb1 + Sb2 = 16 * √53 + 36 + 4 = 40 + 16 * √53 cm2.

Answer: The surface area of ​​the pyramid is 40 + 16 * √53 cm2.



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