A regular triangular prism is split into two prisms by a plane passing through

| September 25, 2021 | education

A regular triangular prism is split into two prisms by a plane passing through the centerlines of the bases. What is the relation of the lateral surfaces of these prisms?

Let the side of the base of the prism be X cm, and the height of the prism is Y cm.

The section formed by the middle lines of the bases of the prism divides it into two prisms.

The RVNKV1M prism is correct.

The middle lines KM and PH are equal to half the lengths of the sides A1C1 and AC, then KM = PH = X / 2 cm.

Since the points K, M, P, H are the midpoints of the sides, the triangles BPH and B1KM are equilateral.

Then Рврн = 3 * РН = 3 * (X / 2) cm.

Then Sbok1 = Pvrn * AA1 = 3 * X * Y / 2 cm2.

An isosceles trapezoid lies at the base of the second prism.

Rarns = X / 2 + X / 2 + X / 2 + X = 5 * X / 2 cm.

Then Sbok2 = Rarns * AA1 = 5 * X * Y / 2 cm2.

The ratio of the lateral surfaces is:

S side2 / S side1 = (5 * X * Y / 2) / (3 * X * Y / 2) = 5/3.

Answer: The area ratio is 5/3.



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