Question

A rectangular metal sheet of length 44 cm and breadth 11 cm is folded along its length to form a cylinder. Find its volume.
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Answer 1

Answer:

Given :-

• A rectangular metal sheet of length 44 cm and breadth 11 cm is folded its length to form a cylinder.

To Find :-

• What is the volume of cylinder.

Solution :-

$\leadsto$ A rectangular metal sheet of length is 44 cm.

So,

$\footnotesize \diamond \: \: \sf\boxed{\bold{Length_{(Rectangular\: Metal\: Sheet)} =\: Circumference_{(Circle)}}}\: \: \: \bigstar$

Given :

• Circumference of Circle = 44 cm

According to the question by using the formula we get,

$\implies \sf\boxed{\bold{Circumference_{(Circle)} =\: 2{\pi}r}}$

where,

• π = Pie or 22/7
• r = Radius

So, by putting those values we get,

$\implies \sf 44 =\: 2 \times \dfrac{22}{7} \times r$

$\implies \sf 44 =\: \dfrac{44}{7} \times r$

$\implies \sf 44 \times \dfrac{7}{44} =\: r$

$\implies \sf \dfrac{308}{44} =\: r$

$\implies \sf 7 =\: r$

$\implies \sf\bold{r =\: 7\: cm}$

Hence, the radius is 7 cm .

☆ Now, we have to find the volume of cylinder :

Given :

• Radius = 7 cm
• Height = 11 cm

According to the question by using the formula we get,

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

where,

• π = Pie or 22/7
• r = Radius
• h = Height

So, by putting those values we get,

$\implies \sf Volume_{(Cylinder)} =\: \dfrac{22}{7} \times (7)^2 \times 11$

$\implies \sf Volume_{(Cylinder)} =\: \dfrac{22}{7} \times (7 \times 7) \times 11$

$\implies \sf Volume_{(Cylinder)} =\: \dfrac{22}{7} \times 49 \times 11$

$\implies \sf Volume_{(Cylinder)} =\: \dfrac{22}{7} \times 539$

$\implies \sf Volume_{(Cylinder)} =\: \dfrac{11858}{7}$

$\implies \sf\bold{\underline{Volume_{(Cylinder)} =\: 1694\: cm^3}}$

$\therefore$ The volume of the cylinder is 1694 cm³ .

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

• A rectangular metal sheet of length 44 cm and breadth 11 cm is folded along its length to form a cylinder.

Let assume that

• Radius of cylinder be r cm.

• Height of cylinder be h cm.

Now, Since a rectangular metal sheet of length 44 cm is folded along its length to form a cylinder.

It means, Circumference of base of cylinder is 44 cm

$\rm \: 2 \: \pi \: r \: = \: 44$

$\rm \: 2 \: \times \: \dfrac{22}{7} \: \times r \: = \: 44$

$\rm \: r \: = \: \dfrac{44 \times 7}{2 \times 22}$

$\bf\implies \:r \: = \: 7 \: cm$

Now, we have to find the volume of cylinder whose radius is 7 cm and height 11 cm.

We know, Volume of cylinder of radius r and height h is given by

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

So, on substituting the values, we get

$\rm \: Volume_{(Cylinder)} \: = \: \dfrac{22}{7} \times 7 \times 7 \times 11$

$\rm \: = \: 22 \times 7 \times 11$

$\rm \: = \: 1694 \: {cm}^{3}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: Volume_{(Cylinder)} = \: 1694 \: {cm}^{3} \: \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$