Question

Find the relation between the volumes and the surface areas of the cylinder, sphere and
cone, when their heights and diameters are equal.

Answer 1

☆Given Question :-

• Find the relation between the volumes and the surface areas of the cylinder, sphere and cone, when their heights and diameters are equal.

☆Given :-

• The height and diameter of cylinder, sphere and cone are equal.

☆To Find :-

• The relationship between the Volumes and Surface Area.

☆Formula used :-

${{ \boxed{\large{\bold\green{Volume_{(Cylinder)}\: = \:\pi r^2 h }}}}}$

${{ \boxed{\large{\bold\pink{Volume_{(Cone)}\: = \:\dfrac{1}{3} \pi r^2 h }}}}}$

${{ \boxed{\large{\bold\purple{Volume_{(Sphere)}\:\ = \:\dfrac{4}{3}\pi r^3}}}}} \:$

☆Solution :-

$\begin{gathered}\begin{gathered}\bf Let = \begin{cases} &\sf{r = radius \: of \: cylinder, \: cone \: and \: sphere} \\ &\sf{h = height \: of \: cylinder, \: cone \: and \: sphere} \end{cases}\end{gathered}\end{gathered}$

☆It is given that height of cylinder, cone and sphere are equal.

☆We know, height of sphere = diameter of sphere.

So, h = 2r

$\sf \: ⟼Volume \: of \: cylinder, V_1 \: = \pi \: {r}^{2} h = 2\pi \: {r}^{3}$

$\sf \: ⟼Volume \: of \: cone, V_3 \: = \dfrac{1}{3} \pi \: {r}^{2} h = \dfrac{2}{3} \pi \: {r}^{3}$

$\sf \: ⟼Volume \: of \: sphere, V_2 \: = \dfrac{4}{3} \pi \: {r}^{3}$

$\sf \: ⟼According \: to \: statement, \: we \: have \: to \: find$

$\sf \: V_1 : V_2 : V_3$

$\sf \: ⟼2\pi \: {r}^{3} : \dfrac{4}{3} \pi \: {r}^{3} : \dfrac{2}{3} \pi \: {r}^{3}$

$\sf \: ⟼6 : 4 : 2$

$\sf \: ⟼3 : 2 : 1$

$\bf\implies \:\ V_1 : V_2 : V_3 = 3: 2 : 1$

☆Case :- 2.

☆Formula used :-

${{ \boxed{\large{\bold\green{Surface Area_{(Cylinder)}\: = \:2\pi r(h +r)}}}}}$

${{ \boxed{\large{\bold\green{Surface Area_{(sphere)}\: = \:4\pi r^2}}}}}$

${{ \boxed{\large{\bold\green{Surface Area_{(Cone)}\: = \:\pi r(l +r)}}}}}$

☆Solution :-

$\sf \: Surface \: Area \: of \: cylinder = 2\pi \: r(h + r)$

$\sf \: ⟼S_1 = 2\pi \: r(2r + r) = 6\pi \: {r}^{2}$

$\sf \: Surface \: Area \: of \: cone \: = \pi \: r(l + r)$

$\sf \: ⟼S_3 = \pi \:r( \sqrt{ {h}^{2} + {r}^{2} } + r)$

$\sf \: ⟼S_3 = \pi \:r( \sqrt{ {(2r)}^{2} + {r}^{2} } + r)$

$\sf \: ⟼S_3 = \pi \:r( \sqrt{ {4r}^{2} + {r}^{2} } + r)$

$\sf \: ⟼S_3 = \pi \:r( \sqrt{ 5 {r}^{2} } + r)$

$\sf \: ⟼S_3 = \pi \:r( \sqrt{5}r + r)$

$\sf \: ⟼S_3 = \pi \: {r}^{2} ( \sqrt{5} +1 )$

$\sf \: Surface \: Area \: of \: sphere,S_2 = 4\pi \: {r}^{2}$

$\sf \: ⟼According \: to \: statement \: we \: have \: to \: find \:$

$\bf \:S_1:S_2:S_3$

$\sf \: ⟼6\pi \: {r}^{2} :4\pi \: {r}^{2} :\pi \: {r}^{2} ( \sqrt{5} +1 )$

$\sf \: ⟼\sf \: ⟼6 : 4 : \sqrt{5} + 1$

$\bf\implies \:S_1:S_2:S_3 = \sf \: ⟼6 : 4 : \sqrt{5} + 1$

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Perimeter of rectangle = 2(length× breadth)

Diagonal of rectangle = √(length ²+breadth ²)

Area of square = side²

Perimeter of square = 4× side

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²