The height of the cylinder is 24 cm, the radius of the base is 4 cm. Determine a) the axial section area.

| August 6, 2021 | education

The height of the cylinder is 24 cm, the radius of the base is 4 cm. Determine a) the axial section area. b) the cross-sectional area of the cylinder is parallel to the axis and 1 cm apart from the axis.

Since the axial section passes through the axis of symmetry of the cylinder, its area will be equal to:

S = 2 * R * h = 2 * 4 * 24 = 192 cm2.

Draw from the center of the circle O segments OA and OB.

Since ОА = ОВ = R = 4 cm, the triangle ОАH is isosceles. The height OH of the triangle divides the base AB into equal segments AH = BH. In a right-angled triangle AOH, we determine the length of the leg AN by the Pythagorean theorem.

AH ^ 2 = OA ^ 2 – OH ^ 2 = 16 – 1 = 15.

AH = √15 cm.

Then AB = 2 * AH = 2 * √15 cm.

Determine the cross-sectional area ABB1A1.

Ssection = AA1 * AB = 24 * 2 * √15 = 48 * √15 cm2.

Answer: The axial cross-sectional area is 192 cm2, the ABB1A1 cross-sectional area is 48 * √15 cm2.



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