Question

the radius and the hight of a cone are in the ratio 4:3. The ratio of the curved surface area to the total surface area of a cone.
(a)5:9
(b)3:7
(c)5:4
(d)16:9​​

Answer 1

$\small \underline \bold{For \: a \: Cone}$-

$\large \underline \bold{Given :-}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$$\sf\pink{\dfrac{r}{h} = \dfrac{4}{3}}$

$\large \underline \bold{To \: Find :-}$

$\: \: \: \: \: \:$$\sf\red{\dfrac{CSA \: of \: Cone}{TSA \: of \: Cone} = \: ?}$

$\large \underline \bold{Using \: Formulas :-}$

$\: \: \: \:$1) $\sf\pink{Curved \: Surface \: Area = \pi r l}$

$\: \: \: \:$2) $\sf\pink{Total \: Surface \: Area = \pi r(r + l)}$

$\: \: \: \:$3) $\sf\pink{Slant \: height =\sqrt{(r^{2} + h^{2})}}$

$\small \bold{Here \: ,}$

$\: \: \: \: \: \: \: \: \:$$\sf{r = radius \: of \: cone}$

$\: \: \: \: \: \: \: \: \:$$\sf{h = perpendicular \: height}$

$\: \: \: \: \: \: \: \: \: \:$$\sf{l = Slant \: height}$

$\large \underline \bold{Solution :-}$

$\: \: \: \: \: \: \: \: \: \:$$\sf{r = 4}$

$\: \: \: \: \: \: \: \: \: \:$$\sf{h = 3}$

$\sf{then \: ,}$

$\: \: \: \: \:$$\sf{l =\sqrt{4^{2} + 3^{2}}}$

$\: \: \: \: \:$$\sf{l =\sqrt{16 + 9} =\sqrt{25}}$

$\: \: \: \: \:$$\sf{l = 5}$

$\sf{Now \: -}$

$\: \: \: \: \: \: \:$$\small \bold{\dfrac{CSA \: of \: Cone}{TSA \: of \: Cone}}$

$\: \: \: \: \:$$\sf{= \dfrac{\cancel{\pi} \cancel{r} l}{\cancel{\pi} \cancel{r} (r + l)}}$

$\: \: \: \: \:$$\sf{= \dfrac{l}{(r + l)}}$

$\: \: \: \: \:$$\sf{= \dfrac{5}{(4 + 5)}}$

$\: \: \: \: \:$$\sf{= \dfrac{5}{9}}$

$\large \underline \bold{Result :-}$

$\: \: \:$$\sf\blue{CSA \: of \: cone \: : \: TSA \: of \: cone = 5 \: : \: 9}$