Answer: (b) = 4.63 cm
. Given,
Height of the cone = AG = 12 cm
Height (h) of the frustum = FG = 4 cm and AF = 8 cm
△AFC ~ △AGE, so
\(\frac{AF}{AG} =\frac{r_1}{r_2}\) ⇒ \(r_1 = \frac{8}{12}r_2 = \frac{2}{3}r_2\)
Volume of frustum = \(\frac{\pi h}{3}(r_1^2+r_2^2+r_1\times r_2)\)
⇒ 190 = \(\frac{22}{7}\times \frac{4}{3}\big(\frac{4}{9}r_2^2+r_2^2+\frac{2}{3}r_2^2\big)\)
⇒ 190 = \(\frac{88}{21}\big(\frac{19}{9}r_2^2\big)\)
⇒ \(r_2^2 = \frac{190\times 9\times 21}{88\times 19}\) = 21.47 (approx)
⇒ r2 = \(\sqrt{21.47}\)
= 4.63 cm (approx.)
