A plane is drawn in the cylinder parallel to the axis and cutting off an arc of 90 degrees from the base circumference.

| January 8, 2021 | education

A plane is drawn in the cylinder parallel to the axis and cutting off an arc of 90 degrees from the base circumference. The diagonal of the section is 10 cm and is 4 cm from the axis. Find the area of the lateral surface of the cylinder.

Since, by condition, the chord AB cuts off the arc, the degree measure of which is 900, the central angle AO1B is also equal to 90. Then the triangle AO1B is isosceles and rectangular.
The height О1Н divides the angle AO1B in half, then the angle AO1H = 45. The angle O1AH = 180 – 90 – 45 = 45.
Then the triangle AO1H is rectangular and isosceles, O1H = AH = 4 cm.
Then AO1 ^ 2 = AH ^ 2 + O1H ^ 2 = 16 + 16 = 32.
AO1 = R = 4 * √2 cm.
Chord length AB = AH * 2 = 4 * 2 = 8 cm.
In a right-angled triangle ABC, according to the Pythagorean theorem, we determine the length of the leg CB.
CB ^ 2 = AC ^ 2 – AB ^ 2 = 100 – 64 = 36.
CB= 6 cm.
Let us determine the area of ​​the lateral surface of the cylinder.
Side = 2 * n * R * CB = 2 * n * 4 * √2 * 6 = 48 * n * √2 cm2.
Answer: The area of ​​the lateral surface of the cylinder is 48 * n * √2 cm2.



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