Two sections perpendicular to the base are drawn through the generatrix of the cylinder

| August 7, 2021 | education

Two sections perpendicular to the base are drawn through the generatrix of the cylinder. cross-sectional areas 12√3 cm2 and 8 cm2 and the angle between them is equal to 150 degrees. Find the area of the axial section of the cylinder.

Let us denote the generatrix of the cylinder through Н, then the cross-sectional areas will be equal:

Sкmm1к1 = Н * КМ = 12 * √3, SmrР1М1 = Н * МР = 8 cm2.

KM = (12 * √3) / N.

MR = 8 / N.

Consider a triangle CMР, in which the angle CMР = 150.

Let’s use the cosine theorem for triangles.

KP ^ 2 = KM ^ 2 + MР ^ 2 – 2 * KM * MР * Cos150 = (12 * √3 / H) ^ 2 + (8 / H) ^ 2 – 2 * (12 * √3 / H) * (8 / H) * (-√3 / 2) = 43 ^ 2 / H ^ 2 + 64 / H ^ 2 + 288 / H ^ 2 = 784 / H ^ 2.

KR = 28 / N.

Consider a triangle CMР inscribed in a circle, through which we determine the radius O1M of the circle.

ОМ = КР / (2 * Sin150).

2 * OM = (28 / H) / (1/2) = 56 / H.

2 * OM = DC, which is the diameter of the circle.

Then the axial section area is S = DC * H = (56 / H) / H = 56 cm2.

Answer: The area of ​​the axial section is 56 cm2.



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