Question

27.
A sphere, a cylinder and a cone are of the same
radius and same height. Find the ratio of their curved
surface.​

Answer 1

$\bold{\underline{\underline{\huge{\sf{AnsWer:}}}}}$

Ratio = 2√2 : √2 : 1

$\bold{\underline{\underline{\large{\sf{StEp\:by\:stEp\:explanation:}}}}}$

GiVeN :

• A sphere, a cylinder and a cone are of the same radius and same height.

To FiNd :

• Ratio of curved surface of sphere, cylinder and cone.

SoLuTiOn :

We have three geometric shapes,

• Sphere
• Cylinder
• Cone

The ratio of curved surface equals to the ratio of the formula of the curved surface.

Sphere :

$\sf{\large{\boxed{\red{4\:\pi\:r^2}}}}$

Cylinder :

$\sf{\large{\boxed{\blue{2\:\pi\:r\:h}}}}$

Cone :

$\sf{\large{\boxed{\red{\pi\:r\:l}}}}$

Given that the shapes are of the same radius and same height.

$\longrightarrow$$\sf{\therefore{r\:=\:h}}$

Now let's move forward with the solution,

$\longrightarrow$$\sf{4\:\pi\:r^2\:\::\:\:2\:\pi\:r\:h\:\::\:\:\pi\:r\:l}$

$\longrightarrow$$\sf{4\:\pi\:r^2\:\::\:\:2\:\pi\:r\:h\:\::\:\:\pi\:r\:({\sqrt{r^2\:+\:h^2)}}}$ $\sf{\because{l\:=\:{\sqrt{r^2\:+\:h^2}}}}$

Now, substitute r instead of h,

$\longrightarrow$$\sf{4\:\pi\:r^2\:\::\:\:2\:\pi\:r\:r\:\::\:\:\pi\:r\:({\sqrt{r^2\:+\:r^2)}}}$

$\longrightarrow$$\sf{4\:\pi\:r^2\:\::\:\:2\:\pi\:r^2\:\::\:\:\pi\:r\:({\sqrt{2\:r^2)}}}$

$\longrightarrow$$\sf{4\:\pi\:r^2\:\::\:\:2\:\pi\:r^2\:\::\:\:\pi\:r\:({\sqrt{2)}\:r}}$

$\longrightarrow$$\sf{4\:\pi\:r^2\:\::\:\:2\:\pi\:r^2\:\::\:\:({\sqrt{2)}\pi\:r\:\times\:r}}$

$\longrightarrow$$\sf{4\:\pi\:r^2\:\::\:\:2\:\pi\:r^2\:\::\:\:({\sqrt{2)}\pi\:r^2}}$

We can now strikeout πr²,

$\longrightarrow$$\sf{4\:\::\:\:2\:\:\::\:\:{\sqrt{2}}}$

Now, we can divide throughout by √2,

$\longrightarrow$$\sf{\dfrac{4}{\sqrt{2}}}$ : $\sf{\dfrac{2}{\sqrt{2}}}$ : $\sf{\dfrac{\sqrt{2}}{\sqrt{2}}}$

Which simplies into,

$\longrightarrow$$\sf{2{\sqrt{2}\:\::\:{\sqrt{2}\:\::\:\:1}}}$