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Question

A right triangle whose sides are 3 cm and 4 cm ( other than hypotenuse ) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. ( Choose the value of $$\pi $$ as found appropriate )

Solution
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i.e $$\dfrac{1}{2}\times OB\times 5=\dfrac{1}{2}\times 4\times 3$$

$$OB=\dfrac{12}{5}=2.4cm$$

Hence $$OB=$$ radius of cone $$=2.4cm$$
Volume of double cone = volume pf cone $$1+$$ volume of cone $$2$$.
$$=\dfrac{1}{3}\pi r^2h_1+\dfrac{1}{3}\pi r^2h_2$$

$$=\dfrac{1}{3}\pi r^2(h_1+h_2)$$

$$=\dfrac{1}{3}\times \pi \times (2.4)^2(OA+AC)$$

$$=\dfrac{1}{3}\times \pi \times 2.4\times 2.4 \times AC$$

$$\dfrac{1}{3}\times \dfrac{22}{7}\times 2.4\times 2.4\times 5$$

$$30.14cm^3$$

Volume of double cone $$=30.14cm^3$$

Surface area of double cone

$$=CSA\,1+CSA\,2$$

$$=\pi rl_1+\pi rl_2=\pi r(l_1+l_2)$$

$$\dfrac{22}{7}\times 2.7\times (4+3)=\dfrac{22}{7}\times 2.7\times 7$$

$$=52.75cm^2$$


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