The radii of the truncated cone bases are 1: 2. The diagonal of the axial section is 15 cm. The height is 12 cm.

| March 9, 2021 | education

The radii of the truncated cone bases are 1: 2. The diagonal of the axial section is 15 cm. The height is 12 cm. Find the volume of the truncated cone.

The volume of the truncated cone is V = 1/3 * pi * H * (r ^ 2 + R * r + R ^ 2).
In the cross-section, an isosceles trapezoid A1AB1B turned out. A1B1 is the diameter of the smaller base of the cone, AB is the diameter of the larger base of the cone.
Triangle А1НВ – rectangular. By the Pythagorean theorem, we find HB: HB = √А1В – А1Н = √225 – 144 = 9 (cm).
AB = 2A1B1, AB = HB + AН,
A1B1 = HB – AH,
HB + AН = 2 (HB – AН),
Empty AH = x. Then:
9 + x = 2 (9-x),
x = 3.
AH = 3 cm,
d1 = HB + AH = 12 cm, so R = 1/2 * d1 = 6 cm.
2 * d2 = d1 = 12,
d2 = 6 cm, so r = 1/2 * d2 = 3 cm.
V = 1/3 * pi * 12 * (36 + 6 * 3 + 9) = 252pi (cc).
Answer: 252pi cc.



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