Question

$\huge \mathscr \color{brown}{\underline{question}} :$
Right circular cylinder and a right circular cone have equal bases and equal heights. if their curved surface are in the ratio of 8:5, show that the radius of their bases is to their height as 3:4 . ​

Answer 1

Solution :

let the radius of cone be r and height, h

It is Given that $\sf \frac{Curved \: surface \: of \: cylinder \: }{Curved \: surface \: of \: cone } = \frac{8}{5}$

$\frac{2πrh}{πrl} = \frac{8}{5} \\ \\ \implies \frac{2h}{l} =\frac{8}{5} \\ \\ \implies l = \frac{5}{4}h$

Now ,

l²=h²+r²

$\implies {\frac{5}{4}h}² = h² + r² \\ \\ \implies 25h²= 16h²+16r² \\ \\ \implies 9h²=16r² \\ \\ \implies \frac{h²}{r²} =\frac{16}{9} \\ \\ \frac{h}{r}=\frac{4}{3} \implies \frac{r}{h}=\frac{3}{4}$

The radius of their bases to their height = 3:4

Answer 2

Answer:

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