In a regular triangular prism ABCDA1B1C1D1, the side of the base is 4 cm. A plane is drawn through

| February 28, 2021 | education

In a regular triangular prism ABCDA1B1C1D1, the side of the base is 4 cm. A plane is drawn through the middle A1C1 and the edge BC. Find the cross-sectional area and the angle of inclination of the secant to the plane of the base. if the lateral surface area is 24 cm 2.

The plane passing through the AC edge and the middle of the A1C1 edge is an isosceles trapezoid НKВС, since KН is the middle line of the A1B1C1 triangle, therefore it is parallel to B1C1 and BC, side ВK = CH.

Then KH = B1C1 / 2 = 4/2 = 2 cm.

Since the triangle at the base is equilateral, the lateral faces of the prism are equal, then Svv1a1a = 24/3 = 8 cm2.

BB1 = Svv1a1a / AB = 8/4 = 2 cm. Then KP = BB1 = 2 cm. BP = AR = AB / 2 = 4/2 = 2 cm.

Then the triangle KРВ is rectangular and isosceles, the angle KВР = 450, ВK = BP / (√2 / 2) = 2 * √2 cm.

Let’s construct a perpendicular KP to the AB side.

In an isosceles trapezoid, we will construct the height of the НM, then the length of the segment BM will be equal to: BM = (BC – KH) / 2 = (4 – 2) / 2 = 1 cm.

In a right-angled triangle ВKM, according to the Pythagorean theorem, KM ^ 2 = ВK ^ 2 – ВM ^ 2 = 8 – 1 = 7.

KM = √7 cm.

Ssection = (BC + KН) * KM / 2 = (4 + 2) * √7 / 2 = 3 * √7 cm2.

Determine the sine of the angle of inclination of the plane. Sinα = KP / KM = 2 / √7.

Angle α = arcsin (2 / √7).

Answer: The cross-sectional area is 3 * √7 cm2, the angle of inclination is arcsin (2 / √7).



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