Given ABCD is the axial section of the cylinder O, the center of the ball circumscribed around the cylinder

| February 11, 2021 | education

Given ABCD is the axial section of the cylinder O, the center of the ball circumscribed around the cylinder Sш = 16п. AB = √15. find the area of the lateral surface of the cylinder.

The axial section of the cylinder is a rectangle AВСD, the diagonal of which is the diameter of the circumscribed ball around the cylinder.

Let us define the radius of the ball through the area of ​​the ball.

Sball = 4 * π * R ^ 2 = 16 * π.

Then R = 2 cm, and therefore, ВD = 2 * R = 2 * 2 = 4 cm.

From the rectangle of the AВD, according to the Pythagorean theorem, we determine the length of the leg of the AD.

AD ^ 2 = DD ^ 2 – AB ^ 2 = 16 – 15 = 1.

AD = 1 cm.

AD is the diameter of the circle at the base of the cylinder.

Then the circumference is equal to: L = π * AD = π cm.

The lateral surface area of ​​the cylinder is equal to: Sside = L * AB = π * √15 cm2.

Answer: The area of ​​the lateral surface of the cylinder is π * √15 cm2.



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