Truncated cone

Find the volume and surface area of the truncated cone if r1 = 12 cm, r2 = 5 cm and side s = 10 cm.

Correct result:

V = 1712.5734 cm³
S = 1064.9999 cm²

Solution:

r_{1} = 12 cm r_{2} = 5 cm s = 10 cm s^{2} = h^{2} + (r_{1} - r_{2})^{2} h = \sqrt{s^{2} - (r_{1} - r_{2})^{2}} = \sqrt{1 0^{2} - (12 - 5)^{2}} = \sqrt{51} cm ≐ 7.1414 cm V = \frac{π \cdot h}{3} \cdot (r_{1}^{2} + r_{1} \cdot r_{2} + r_{2}^{2}) = \frac{3.1416 \cdot 7.1414}{3} \cdot (1 2^{2} + 12 \cdot 5 + 5^{2}) = 1712.5734 {cm}^{3}

S = π \cdot r_{1}^{2} + π \cdot r_{2}^{2} + π \cdot (r_{1} + r_{2}) \cdot \sqrt{h^{2} + (r_{1} - r_{2})^{2}} = 3.1416 \cdot 1 2^{2} + 3.1416 \cdot 5^{2} + 3.1416 \cdot (12 + 5) \cdot \sqrt{7.141 4^{2} + (12 - 5)^{2}} = 1064.9999 {cm}^{2}

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