In an inclined triangular prism, the areas of the two faces are 15 and 25 cm ^ 2.

| April 26, 2021 | education

In an inclined triangular prism, the areas of the two faces are 15 and 25 cm ^ 2. The angle between them is 120 °. Find the area of the lateral surface of the prism if the length of the lateral rib is 5 cm.

From the vertex A1 of the prism, construct perpendiculars A1M to the edge CC1 and A1K to the edge BB1.

Since, by condition, the side faces АА1С1С and СС1В1В are perpendicular, the triangle А1МК is rectangular.

Let the area of ​​the side face AA1B1B be equal to 25 cm2, and the area of ​​the face CC1B1B equal to 15 cm2.

Determine the heights A1M and MK of the side faces.

A1M = Saa1c1c / CC1 = 25/5 = 5 cm.

MK = Scc1v1v / CC1 = 15/5 = 3 cm.

Since the angle between the side faces is 1200, then in the A1MK triangle, the angle A1MK = 1200, then, by the cosine theorem, we determine the length of the A1K side.

A1K ^ 2 = A1M ^ 2 + MK ^ 2 – 2 * A1M * MK * Cos120 = 25 + 9 – 2 * 5 * 3 * (-1/2) = 34 + 15 = 49.

A1K = 7 cm.

Then Pa1mk = 3 + 5 + 7 = 15 cm.

Sside = Pa1mk * CC1 = 15 * 5 = 75 cm2.

Answer: The lateral surface area is 75 cm2.



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