Question

10. If the radius of a cylinder is 14 cm and the height is 5 cm, then find the volume of
this cylinder. (=22/7)​

Answer 1

Answer:

Volume of the cylinder = 3080 cm³

Step-by-step explanation:

Given:

• Radius of the cylinder = 14 cm
• Height of the cylinder = 5 cm

To find:

• Volume of the cylinder.

Solution:

• Radius (r) = 14 cm
• Height (h) = 5 cm

We know that,

${\boxed{\sf{Volume\:of\: cylinder=\pi\:r^2h}}}$

• [put values]

$\implies\sf{Volume\:of\: the\: cylinder=\dfrac{22}{7}\times\:(14)^2\times\:5\:cm^3}$

$\implies\sf{Volume\:of\: the\: cylinder=\dfrac{22}{7}\times\:14\times\:14\times\:5\:cm^3}$

$\implies\sf{Volume\:of\: the\: cylinder=22\times\:2\times\:14\times\:5\:cm^3}$

$\implies\sf{Volume\:of\: the\: cylinder=3080\:cm^3}$

Therefore, the volume of the cylinder is 3080 cm³.

________________

★ Additional Info :

• 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

Answer 2

$\large\underline{ \underline{ \sf \maltese\red{ \: Question:- }}}$

• If the radius of a cylinder is 14 cm and the height is 5 cm, then find the volume of the cylinder. Take (π=22/7)

$\large\underline{ \underline{ \sf \maltese\red{ \: Given:- }}}$

• $\sf\green{Radius \: of \: Cyclinder \: = \: \blue{\fbox\orange{14cm }}}$
• $\sf\green{Height\: of \: Cyclinder \: = \: \blue{\fbox\orange{5cm }}}$

$\large\underline{ \underline{ \sf \maltese\red{ \: Diagram:- }}}$

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\multiput(-0.5,-1)(26,0){2}{\line(0,1){40}}\multiput(12.5,-1)(0,3.2){13}{\line(0,1){1.6}}\multiput(12.5,-1)(0,40){2}{\multiput(0,0)(2,0){7}{\line(1,0){1}}}\multiput(0,0)(0,40){2}{\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\multiput(18,2)(0,32){2}{\sf{14cm}}\put(9,17.5){\sf{5cm}}\end{picture}$

━━━━━━━━━━━━━━━━━━━━━━━━━

$\large\underline{ \underline{ \sf \maltese\red{ \: Solution:- }}}$

$\begin{gathered}\begin{gathered}\begin{gathered}: \implies \underline\blue{ \boxed{\displaystyle \sf \bold\orange{\: Volume \: of \: Cyclinder \: = \:\pi \: {r}^{2} h } }}\\ \\\end{gathered}\end{gathered}\end{gathered}$

$\qquad \quad {:} \longrightarrow \sf{\sf{Volume \: = \: \frac{22}{7} \: \times \: (14 )^{2} \: \times \: 5 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{Volume \: = \: \frac{22}{7} \: \times \: 14 \: \times \: 14 \: \times \: 5 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{Volume \: = \: \frac{22}{ \cancel7} \: \times \: \cancel{14} \: \times \: 14 \: \times \: 5 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{Volume \: = \: {22} \: \times \: 2 \: \times \: 14 \: \times \: 5 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{Volume \:of \: cyclinder = \: 3080 {cm}^{3} }}$

$\qquad\quad {:} \longrightarrow \underline \red{\boxed{\sf{Volume \: of \: Cyclinder \: = \: 3080 {cm}^{3} }}}$

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More about Volume :-

• A solid figure occupies space. The magnitude of the space occupied by a solid figure is called its volume.
• The volume of an object / solid refer to the space occupied by the object/ solid.
• The capacity of a container refers to the quantity that a container can hold.
• If the volumes of any two solid figures are equal then their capacities or space occupied are also equal . , However their surface areas may be different

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