In a regular triangular truncated pyramid, the lateral surface area is 720 cm2

univerkov | April 24, 2021 | education

In a regular triangular truncated pyramid, the lateral surface area is 720 cm2, and the sides of the base are 8 and 24 cm. Determine the length of the lateral edge of the pyramid.

The lateral faces of a regular truncated triangular pyramid are equal-sized isosceles trapezoids. Then the area of the trapezoid ABCD is equal to: S = S side / 3 = 720/3 = 240 cm2.

Knowing the area, we determine the height of the ВН of the side face.

S = (AD + BC) * ВН / 2.

BH = 2 * S / (AD + BC) = 2 * 240/32 = 15 cm.

Since the side face of ABCD is an isosceles trapezoid, then AH = (AD – BC) / 2 = (24 – 8) / 2 = 8 cm.

In a right-angled triangle ABН, AB ^ 2 = AH ^ 2 + BH ^ 2 = 64 + 225 = 289. AB = 17 cm.

Answer: The length of the side edge is 17 cm.

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