Question

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

Answer 1

Answer:

Since, the height of a sphere is the diameter, the cone and cylinder have height 2r.

Then

Curved surface area of Sphere=4πr²

Curved surface area of cylinder =2πr(2r)=4πr²

Curved surface area of cone = πrl

where, l=

(r

2

+h

2

)=

(r2+(2r)

2

)=

5r

2

=r

5

⇒ Curved surface area of cone =π

5r

2

Now,

ratio of CSA 's a sphere ,cylinder and a cone = 4πr

2

:2πrh:πrl

=4πr

2

:4πr

2

:πr

2

5

Answer 2

$\implies\huge \sf \underline \green{Answer}$

$\huge\implies\rm{ \underline{ \underline{ \purple{ \rm{4 : 4 : \sqrt{5} \: }}}} }$

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$\implies \huge \sf \underline \red{Q uestion}$

• A sphere , a cylinder and a cone are of same radius and same height .Find the ratio of their curved surfaces.

________________________________________________

$\implies\sf \huge \underline \pink{To \: find}$

• ratio of curved surfaces areas

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$\implies \sf \huge \underline \orange{solution}$

• height of a sphere is diameter

• height of a cone is 2r

• height of cylinder is 2r

• ratio of curved surfaces is ?

Do you know that formula of curved surface area of sphere,cone and cylinder is

• csa is curved surface area

____________________________________________________

$\: \: \: \: \: \: \: \sf{ \boxed{ \underline{ \underline{ \blue{ \sf{CSA \: of \: sphere = 4\pi {r}^{2}}}}}}}$

$\: \: \: \: \: \: \sf{ \boxed{ \underline{ \underline{ \purple{ \sf{CSA \: of \: cylinder = 2\pi \: r(2r) = 4\pi{r}^{2}}}}}}}$

$\: \: \: \: \: \: \: \sf{ \boxed{ \underline{ \underline{ \pink{ \sf{CSA \: of \: cone = \pi \: rl}}}}}}$

• so, find the csa of cone

$\rm \red{1 = \sqrt{( {r}^{2} + {h}^{2}) } - \sqrt{(r2 + (2r) {}^{2}) } - \sqrt{5 {r}^{2} - r \sqrt{5}} }$

$\: \: \: \: \: \: \: \sf{ \boxed{ \underline{ \underline{ \pink{ \sf{CSA \: of \: cone = \pi \: \sqrt{5 {r}^{2} } }}}}}}$

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• Now we have to find the ratio of curved surface

Ratio of CSA of sphere, cylinder and cone is

$\tt \huge{ \boxed{ \underline{ \underline{ \green{ \tt{4\pi \: {r}^{2} :2\pi \: rh : \: \pi \: rl}}}}}}$

$\implies\rm \red{4\pi \: {r}^{2} :2\pi \: rh : \: \pi \: rl}$

$\implies\rm \red{4\pi \: {r}^{2} :4\pi \: {r}^{2}: \pi \: {r}^{2} \sqrt{5} }$

$\implies\rm \red{4 : 4 : \sqrt{5} }$