The sides of the bases of a regular triangular truncated pyramid are 4 cm and 8 cm.

| September 25, 2021 | education

The sides of the bases of a regular triangular truncated pyramid are 4 cm and 8 cm. The height of the truncated pyramid is √10 cm. Find the area of the lateral surface of the truncated pyramid.

Let’s draw the heights BH and B1H1 of equilateral triangles at the base of the pyramid and determine their lengths.

h = a * √3 / 2, where a is the side of the triangle.

BH = 8 * √3 / 2 = 4 * √3 cm.

B1H1 = 4 * √3 / 2 = 2 * √3 cm.

The intersection point of the heights O and O1 divides the height in a ratio of 2/1.

Then OH = BH / 3 = 4 * √3 / 3 cm.

O1H1 = B1H1 / 3 = 2 * √3 / 3 cm.

Then the segment НР = ОН – О1Н1 = 4 * √3 / 3 – 2 * √3 / 3 = 2 * √3 / 3 cm.

From the right-angled triangle НН1Р we find the apothem НН1.

НН1 ^ 2 = Н1Р ^ 2 + НР ^ 2 = (√10) ^ 2 + (2 * √3 / 3) ^ 2 = 10 + 4/3 = 34/3.

НН1 = √ (34/3) cm.

Let’s find the area of ​​the lateral surface.

Side = 3 * (АС + А1С1) * НН1 / 2 = 3 * 12 * √ (34/3) / 2 = 18 * √34 / √3 = 18 * √3 * √34 / √3 * √3 = 6 * √102 cm2.

Answer: The lateral surface area is 6 * √102 cm2.



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