Find the area of the lateral surface of a regular triangular truncated pyramid

| September 16, 2021 | education

Find the area of the lateral surface of a regular triangular truncated pyramid, the sides of the base, which are 3 and 11, and the lateral edge 5.

Since the lateral edges of the truncated pyramid are 5 cm, its lateral faces are isosceles trapezoids with bases of 3 cm and 11 cm.

Let’s build the height А1Н of the trapezoid АА1С1С.

The height of an isosceles trapezoid divides the larger base into two segments, the length of the smaller of which is equal to the half-difference of the lengths of the bases.

AH = (AC – A1C1) / 2 = (11 – 3) / 2 = 4 cm.

In a right-angled triangle АА1Н, according to the Pythagorean theorem, А1Н ^ 2 = АА1 ^ 2 – АН ^ 2 = 25 – 16 = 9. А1Н = 3 cm.

Then Saa1c1c = (A1C1 + AC) * A1H / 2 = (3 * 11) * 3/2 = 21 cm2.

Side = 3 * Saa1s1s = 3 * 21 = 63 cm2.

Answer: The lateral surface area is 63 cm2.



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