$A$ fez,the cap used by the Turks,is shaped like the frustum of a cone (see $Fig.$). If its radius on the open side is $10 \, cm$,radius at the upper base is $4 \, cm$ and its slant height is $15 \, cm$,find the area of material used for making it.
(N/A) Radius $(r_2)$ at the upper circular end $= 4 \, cm$.
Radius $(r_1)$ at the lower circular end $= 10 \, cm$.
Slant height $(l)$ of the frustum $= 15 \, cm$.
Area of material used for making the fez $=$ $CSA$ of the frustum $+$ Area of the upper circular end.
Area $= \pi(r_1 + r_2)l + \pi r_2^2$.
Area $= \pi(10 + 4) \times 15 + \pi(4)^2$.
Area $= \pi(14) \times 15 + 16\pi$.
Area $= 210\pi + 16\pi = 226\pi$.
Using $\pi = \frac{22}{7}$,Area $= 226 \times \frac{22}{7} = \frac{4972}{7} \, cm^2$.
Area $= 710 \frac{2}{7} \, cm^2$.
Therefore,the area of material used for making it is $710 \frac{2}{7} \, cm^2$.
Radius $(r_1)$ at the lower circular end $= 10 \, cm$.
Slant height $(l)$ of the frustum $= 15 \, cm$.
Area of material used for making the fez $=$ $CSA$ of the frustum $+$ Area of the upper circular end.
Area $= \pi(r_1 + r_2)l + \pi r_2^2$.
Area $= \pi(10 + 4) \times 15 + \pi(4)^2$.
Area $= \pi(14) \times 15 + 16\pi$.
Area $= 210\pi + 16\pi = 226\pi$.
Using $\pi = \frac{22}{7}$,Area $= 226 \times \frac{22}{7} = \frac{4972}{7} \, cm^2$.
Area $= 710 \frac{2}{7} \, cm^2$.
Therefore,the area of material used for making it is $710 \frac{2}{7} \, cm^2$.
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