Question

A right circular cylinder and a cone have equal bases and equal heights. If their curved surface areas are in the ratio 8:5, show that the ratio between radius of theirbases to their height is 3:4

Answer 1

Let the radius and height of cylinder and cone be r and h.

Let the slant height of cone be l.

CSA of cylinder : 2πrh

CSA of cone : πrl

Ratio = CSA of cylinder/CSA of cone = 2πrh/πrl = 8/5

⇒ h/l = 4/5

⇒ h²/l² = 16/25

⇒l² = (25/16)h²

⇒ h² + r² = (25/16)h²

⇒ r² = (9/16)h²

⇒ (r/h)² = (3/4)²

⇒ r/h = 3/4

hence ratio of radius to height is 3:4

Hence Proved

Answer 2

Question:

→ A right circular cylinder and a cone have equal bases and equal heights. If their curved surface areas are in the ratio 8:5, show that the ratio between radius of theirbases to their height is 3:4 .

Step-by-step explanation:

Given :-

→ Ratio of CSA of right circular cone and cone is 8 : 5 .

To Prove :-

→ The ratio between radius of their bases to their height is 3 : 4 .

Solution :-

∵ CSA of cylinder = 2πrh .

∵ CSA of cone = πrl .

Now,

$\sf \implies \frac{2 \cancel{ \pi r}h }{ \cancel{ \pi r}l } = \frac{8}{5} .\\ \\ \sf \implies \frac{2h}{l} = \frac{8}{5} .\\ \\ \sf \implies \frac{2h}{ \sqrt{h^2 + r^2 }} = \frac{8}{5} . \\ \\ \sf \implies { \big( \frac{2h}{ \sqrt{h^2 + r^2 }} } \big) ^2= ( \frac{8}{5} )^2 . \bigg[ Squaring \: both \: side . \bigg] \\ \\ \sf \implies \frac{4h^2}{h^2 + r^2 } = \frac{64}{25} . \\ \\ \sf \implies 100h^2 = 64h^2 + 64r^2 .\\ \\ \sf \implies 100h^2 - 64h^2 = 64r^2 .\\ \\ \sf \implies 36h^2 = 64r^2 .$

$\sf \implies \frac{36}{64} = \frac{r^2}{h^2} .\\ \\ \sf \implies \frac{r}{h} = \sqrt{\frac{36}{64} } .\\ \\ \sf \implies \frac{r}{h} = \frac{6}{8} = \frac{3}{4} .\\ \\ \sf \implies \frac{r}{h} = \frac{3}{4} .\\ \\ \huge \boxed {\boxed{ \pink{ \sf \therefore r : h = 3 : 4 . }}}$

Hence, it is solved .

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