In a regular triangular truncated pyramid, the sides of the bases 5 and 7, and the lateral rib are inclined at an angle

| August 5, 2021 | education

In a regular triangular truncated pyramid, the sides of the bases 5 and 7, and the lateral rib are inclined at an angle of 45 to the base. Find the side surface of the pyramid.

Let us determine their heights in the equilateral triangles lying at the bases of the pyramid.

AH = BC * √3 / 2 = 7 * √3 / 2 cm.

A1H1 = B1C1 * √3 / 2 = 5 * √3 / 2 cm.

The heights of the triangles are their medians, and at the points of intersection they are divided in a ratio of 2/2.

Then O1H1 = A1H1 / 3 = 5 * √3 / 6 cm.

OH = AH / 3 = 7 * √3 / 6.

Quadrangle OO1N1N is a rectangular trapezoid in which we draw the height KN1. Then KH = OH – OK = OH – O1H1 = 7 * √3 / 6 – 5 * √3 / 6 = √3 / 3 cm.

Let’s build the height A1M. Triangle AA1M is rectangular and isosceles, AM = A1M.

AM = AH – A1H1 – KH = 7 * √3 / 2 – 5 * √3 / 2 – √3 / 3 = 4 * √3 / 6 = 2 * √3 / 3 cm.

Then A1M = H1K = 2 * √3 / 3.

In a right-angled triangle КНН1, according to the Pythagorean theorem, НН12 = Н1К2 + КН2 = (2 * √3 / 3) 2 + (√3 / 3) 2 = 5/3.

НН1 = √ (5/3) cm.

The side face of the prism is a trapezoid, then Svss1v1 = (B1C1 + BC) * HH1 / 2 = 12 * √ (5/3) / 2 = 6 * √ (5/3) = 2 * √15 cm2.

Then Side = 3 * Svss1v1 = 3 * 2 * √15 = 6 * √15 cm2.

Answer: The lateral surface area is 6 * √15 cm2.



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