In a triangular truncated pyramid with a height equal to 10, the sides of one of the bases are 27,29,52

In a triangular truncated pyramid with a height equal to 10, the sides of one of the bases are 27,29,52. Determine the volume of the truncated pyramid if the perimeter of the other base is 72.

Let’s define the perimeter of the triangle ABC.

Ravs = (27 + 29 + 52) = 108 cm.

Triangles ABC and A1B1C1 are similar, and the ratio of their perimeters is equal to the coefficient of their similarity. K = Pa1v1s1 / Ravs = 72/108 = 2/3.

By Heron’s theorem, we determine the area of the triangle ABC.

The semi-perimeter of the A1B1C1 triangle is: p = 108/2 = 54 cm.

Sa1v1s1 = √54 * (54 – 27) * (54 – 29) * (54 – 52) = √72900 = 270 cm2.

Sa1b1s1 / Saabs = K2 = 4/9.

Then Sa1v1s1 = Savs * 4/9 = 270 * 4/9 = 120 cm2.

Let’s define the volume of the truncated pyramid.

V = OO1 * (Savs + Sa1v1s1 + √Savs * Sa1v1s1) / 3 = 10 * (270 + 120 + √32400) / 3 = 1900 cm3.

Answer: The volume of the pyramid is 1900 cm3.



One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.