The axial section of the cylinder is a rectangle, the side of which coincides with the diameter

The axial section of the cylinder is a rectangle, the side of which coincides with the diameter of the cylinder, K times smaller than its diagonal. Find the ratio of the lateral surface of the cylinder to the area of its base

Let the height CD of the axial section ABCD = X cm, then, by condition, its diagonal is equal to K * X cm.

The triangle ACD is rectangular, then, by the Pythagorean theorem, AD ^ 2 = D ^ 2 = AC ^ 2 – CD ^ 2 = K ^ 2 * X ^ 2 – X ^ 2 = X ^ 2 * (K ^ 2 – 1).

AD = X * √ (K ^ 2 – 1).

Let us determine the area of the lateral surface of the cylinder.

Sside = π * AD * CD = π * X * √ (K ^ 2 – 1) * X = π * X ^ 2 * √ (K ^ 2 – 1).

Determine the area of the base of the cylinder.

Sop = π * AD ^ 2/4 = π * (X * √ (K ^ 2 – 1)) ^ 2/4.

Then S side / S main = π * X ^ 2 * √ (K ^ 2 – 1) / (π * (X * √ (K ^ 2 – 1)) ^ 2/4) = 4 * / √ (K2 – 1) …

Answer: The area ratio is 4 * / √ (K ^ 2 – 1).