The diagonal of the section of the cylinder, parallel to the axis, is 6 cm and makes an angle of 45 degrees with the plane of the lower base. This section cuts off a 60-degree arc at the base. Find the area of the lateral surface of the cylinder.
Consider a right-angled triangle ACD, in which, by condition, the hypotenuse AC = 6 cm, and the angle CAD = 45, then the angle ACD = 180 – 90 – 45 = 45.
Triangle ACD is isosceles, AD = CD.
AD = AC = AC * Sin45 = 6 * √2 / 2 = 3 * √2 cm.
According to the condition, the degree measure of the arc AD = 60, then the value of the central angle AOD is also equal to 60. Since in the triangle AOD AO = OD, the triangle AOD is equilateral AD = OA = OD.
OA is the radius of the base of the cylinder, and AB is its height.
Let us determine the area of the lateral surface of the cylinder.
Side = 2 * n * OA * AB = 2 * n * 3 * √2 * 3 * √2 = 36 * n cm2.
Answer: Side = 36 * n cm2.
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