Question 14 Marks
The radii of the circular ends of a frustum of a cone are 14 cm and 8 cm. If the height of the frustum is 8 cm, find : ($\pi=3.14$)
(i) Curved surface area of frustum.
(ii) Total surface area of frustum.
(iii) Volume of the frustum.
(i) Curved surface area of frustum.
(ii) Total surface area of frustum.
(iii) Volume of the frustum.
Answer
View full question & answer→$R=14, r=8, h=8$.
Slant height $l = \sqrt{h^{2} + (R-r)^{2}} = \sqrt{8^{2} + 6^{2}} = 10$ cm.
(i) CSA = $\pi(R+r)l = 3.14(14+8)10 = 3.14 \times 220 = 690.8$ sq.cm.
(ii) TSA = CSA + $\pi R^{2} + \pi r^{2} = 690.8 + 3.14(196 + 64) = 690.8 + 816.4 = 1507.2$ sq.cm.
(iii) Volume = $\frac{1}{3}\pi h(R^{2} + r^{2} + Rr) = \frac{1}{3} \times 3.14 \times 8 \times (196 + 64 + 112) = \frac{1}{3} \times 25.12 \times 372 = 3114.88$ cm$^{3}$.
Slant height $l = \sqrt{h^{2} + (R-r)^{2}} = \sqrt{8^{2} + 6^{2}} = 10$ cm.
(i) CSA = $\pi(R+r)l = 3.14(14+8)10 = 3.14 \times 220 = 690.8$ sq.cm.
(ii) TSA = CSA + $\pi R^{2} + \pi r^{2} = 690.8 + 3.14(196 + 64) = 690.8 + 816.4 = 1507.2$ sq.cm.
(iii) Volume = $\frac{1}{3}\pi h(R^{2} + r^{2} + Rr) = \frac{1}{3} \times 3.14 \times 8 \times (196 + 64 + 112) = \frac{1}{3} \times 25.12 \times 372 = 3114.88$ cm$^{3}$.