a right triangle whose sides are 6cm and 8 cm ( other than hypotenuse ) is made to revolve about - Maths - Surface Areas and Volumes

Question

a right triangle whose sides are 6cm and 8 cm ( other than
hypotenuse ) is made to revolve about its hypotenuse find the
volume and surface area of the double cone so formed

Koka Sri Lakshmi Divya Sai · Accepted Answer

$The double cone so formed the right angled triangle ABC about its hypotenuse is shown in fig hypotenuse AC= \sqrt{6^{2} + 8^{2}} = \sqrt{36 + 64} = \sqrt{100} = 10 c m Area of Δ ABC= \frac{1}{2} \times AB \times AC \frac{1}{2} \times AC \times OB = \frac{1}{2} \times 6 \times 8 \Rightarrow \frac{1}{2} \times 10 \times OB=24 \Rightarrow OB=4.8cm Volume of double cone=volume of cone1+volume of cone2 = \frac{1}{3} π r^{2} h_{1} + \frac{1}{3} π r^{2} h_{2} = \frac{1}{3} π r^{2} (h_{1} + h_{2}) = \frac{1}{3} π r^{2} (OA+OC) = \frac{1}{3} \times \frac{22}{7} \times (4.8)^{2} \times 10 = 241.3 c m^{3} Surface area of double cone=surface area of cone1+surface area of cone2 = π r l_{1} + π r l_{2} = π r (6 + 8) = \frac{22}{7} \times 4.8 \times 14 = 211.2 c m^{2}$

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a right triangle whose sides are 6cm and 8 cm ( other than hypotenuse ) is made to revolve about its hypotenuse. find the volume and surface area of the double cone so formed.