Question

the radius of the bases of a cylinder and a cone are in the ratio 3 ratio 4 and their Heights are in the ratio 2 ratio 3 then find the ratio in their volume

Answer 1

Hey Mate
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Here is your Answer
.Let the radii of the bases of a cylinder and a cone be 3r and 4r.

Let the heights of the bases of a cylinder and a cone be 2h and 3h.

The volume of the cylinder =(pi)(3r)^2(2h) = 18(pi)r^2*h.

The volume of the cone=(pi)(4r)^2(3h)/3 = (16)(pi)r^2*h.

The ratio of the volumes of the cylinder to that of the cone = 18:16 or 9:8.

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Answer 2

Question:

→ The radius of the bases of a cylinder and a cone are in the ratio 3 : 4 and their Heights are in the ratio 2 : 3 then find the ratio in their volume .

Answer:

→ Ratio in their volume = 9 : 8 .

Step-by-step explanation:

Given :-

→ r₁ : r₂ = 3 : 4 .

→ h₁ : h₂ = 2 : 3 .

To find :-

→ V₁ : V₂ .

Solution :-

∵ Volume of cylinder ( V₁ ) = πr₁²h₁ .

∵ Volume of cone (V₂ ) = 1/3 πr₂²h₂ .

$\sf \because \frac{V_1}{V_2} .\\ \\ \sf = \frac{ \cancel\pi {r_1} ^2 h_1 }{ \frac{1}{3} \cancel\pi { r_2}^2 h_2 } .\\ \\ \sf = 3 ( \frac{r_1}{r_2} )^2 \times ( \frac{h_1}{h_2}) .\\ \\ \sf = \cancel3 ( \frac{3}{4})^2 \times \frac{2}{\cancel3} .\\ \\ \sf = \frac{9}{16} \times 2 .\\ \\ \sf \implies \frac{V_1}{V_2} = \frac{9}{8} .\\ \\ \huge \boxed{\boxed{ \green { \sf \therefore V_1 : V_2 = 9 : 8 .}}}$

Hence, it is solved ,

THANKS .