The surface

The surface of a truncated rotating cone with side s = 13 cm is S = 510π cm². Find the radii of the bases when their difference in lengths is 10cm.

Correct answer:

r₁ = 15 cm
r₂ = 5 cm

s = 13 cm S = 510 π = 510 \cdot 3.1416 ≐ 1602.2123 cm^{2} S = π \cdot r_{1}^{2} + π \cdot r_{2}^{2} + π \cdot (r_{1} + r_{2}) \cdot s r_{1} = r_{2} + 10 S = π \cdot r_{1}^{2} + π \cdot (r_{1} - 10)^{2} + π \cdot (r_{1} + (r_{1} - 10)) \cdot s S = π \cdot x^{2} + π \cdot (x - 10)^{2} + π \cdot (x + (x - 10)) \cdot s 1602.2122533308 = π \cdot x^{2} + π \cdot (x - 10)^{2} + π \cdot (x + (x - 10)) \cdot 13 - 6.2831853071796 x^{2} - 18.85 x + 1696.46 = 0 6.2831853071796 x^{2} + 18.85 x - 1696.46 = 0 a = 6.283185; b = 18.85; c = - 1696.46 D = b^{2} - 4 a c = 18.8 5^{2} - 4 \cdot 6.283185 \cdot (- 1696.46) = 42991.996771145 D > 0 x_{1, 2} = \frac{- b \pm D}{2 a} = \frac{- 1 8 . 8 5 \pm 4 2 9 9 2}{1 2 . 5 6 6 3 7 1} x_{1, 2} = - 1.5 \pm 16.5 x_{1} = 15 x_{2} = - 18 r_{1}, r_{2} > 0 r_{1} = x_{1} = 15 = 15 cm

r_{2} = r_{1} - 10 = 15 - 10 = 5 = 5 cm Verifying Solution: S_{2} = π \cdot r_{1}^{2} + π \cdot r_{2}^{2} + π \cdot (r_{1} + r_{2}) \cdot s = 3.1416 \cdot 1 5^{2} + 3.1416 \cdot 5^{2} + 3.1416 \cdot (15 + 5) \cdot 13 ≐ 1602.2123 cm^{2} S_{2} = S

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