Question

12. A rectangular piece of paper having
length 4.4 cm and breadth 1.5 cm is to
be rolled along its length to form a right
circular cylinder. Find the height and the
volume of the right circular cylinder.
n.
Trap Up
​

Answer 1

Given -

• Length of rectangle = 4.4 cm

• Breadth of rectangle = 1.5cm

To find -

• The height and volume of Cylinder formed.

Formula used -

• Circumference of circle.

• Volume of cylinder.

Solution -

In the question, we are provided with the dimensions of rectangle, And then, it was rolled from the length side to form a cylinder, and we need to find the height and the volume of the cylinder. First we will find the radius of the cylinder by using the circumference of circle formula and then we will find the volume of the cylinder, by using volume of cylinder formula. Let's do it !

So -

As it is given that, the rectangle is rolled from length side of the rectangle, so, the height of the cylinder will be equal to the Length of the cylinder.

$\sf \: length \: of \: rectangle \: = height \: of \: cylinder$

$\sf \: 4.4cm \: = height \: of \: cylinder$

$\therefore$Height of cylinder is 4.4 cm

Now -

For finding the volume of cylinder, we need radius. As the length of the rectangle is equal to the height of cylinder, then the breadth of the rectangle will be equal to the circumference of the cylinder. And from the formula of circumference of the circle, we will find the radius. Here, we will take, π as 3.14!

$\sf \: circumfrence = breadth \: of \: rectangle$

$\sf \longrightarrow\: 2 \: \pi \: r \: = 1.5 \: cm$

$\sf \longrightarrow \: 2 \: \times \: 3.14 \: \times r \: = 1.5 \:$

$\sf \longrightarrow \: 6.28 \: \times r = 1.5$

$\sf \longrightarrow \: r \: = \dfrac{1.5}{6.28}$

$\sf \longrightarrow \: r \: = 0.238 \: cm$

Now -

We have height and we have radius so, now, we will apply the formula of volume of cylinder, to obtain the same.

$\sf \underline{volume \: of \: cylinder} \: = \pi \: {r}^{2} \: h$

On substituting the values -

$\sf \: volume \: = 3.14 \: \times {(0.238)}^{2} \times 4.4 \: cm$

$\sf \: volume \: = 3.14 \: \times 0.0566 \: \times 4.4 \: cm$

$\sf \: volume \: = 0.1778 \: \times 4.4 \: cm$

$\sf \: volume \: = 0.782 \: {cm}^{3}$

$\therefore$ The height of cylinder is 4.4 cm and volume is 0.782 cm³

_____________________________

Answer 2

${\large{\bold{\rm{\underline{Understanding \; the \; question}}}}}$

✨ This question says that a rectangular piece of paper having length as 4.4 cm and breadth as 1.5 cm and it is to be rolled along it's length to form a right circular cylinder. We have to find the height and the volume of the right circular cylinder.

${\large{\bold{\rm{\underline{Given \; that}}}}}$

✨ Length of rectangular piece of paper = 4.4 cm

✨ Breadth of rectangular piece of paper = 1.5 cm

✨ Rectangular piece of paper is rolled along it's length to form a right circular cylinder

${\large{\bold{\rm{\underline{To \; find}}}}}$

✨ The height of the right circular cylinder.

✨ The volume of the right circular cylinder.

${\large{\bold{\rm{\underline{Solution}}}}}$

✨ The height of the right circular cylinder = 4.4 cm

✨ The volume of the right circular cylinder = 0.782 cm³

${\large{\bold{\rm{\underline{Using \; concepts}}}}}$

✨ Formula to find volume of cylinder.

✨ Formula to find circumference of circle

${\large{\bold{\rm{\underline{Using \; formula}}}}}$

✨ Volume of cylinder = πr²h

✨ Circumference of circle = 2πr

$\; \; \; \; \; \; \; \; \; \;{\bf{Where,}}$

● r denotes radius

● π is pronounced as pi

● Value of π is 22/7 or 3.14

● h denotes height

● ² means square

${\large{\bold{\rm{\underline{Full \; Solution}}}}}$

~ We have to find the height and the volume of the right circular cylinder.

~ So we have to use formula to find volume of cylinder and circumference of a circle too.

~ As in the question it's given that rectangular piece of paper is to be rolled along it's length to form a right circular cylinder. So, length of rectangular piece of paper is equal to height of the cylinder here !..

• Length of rectangular piece = Height of cylinder.

• Length of rectangle = Height of cylinder.

• 4.4 cm = Height of cylinder.

Therefore, height of cylinder = 4.4 cm

~ Now according to the question, its already cleared that rectangular piece of paper is to be rolled along it's length to form a right circular cylinder. So, length of rectangular piece of paper is equal to height of the cylinder !. Henceforth, breadth of rectangular piece of paper is equal to the cylinder's circumference..!

• Circumference = Breadth of rectangular piece

• 2πr = 1.5 cm

Henceforth,

➝ 2 × 22/7 × r = 15

➝ 44/7 × r = 15

• Cross multiplying (÷ = ×) ; (× = ÷)

➝ r = 0.238 cm

~ Now let's find the by using the formula to find volume of cylinder !..

➝ Volume of cylinder = πr²h

➝ Volume of cylinder = 22/7 × 0.238² × 4.4

➝ Volume of cylinder = 22/7 × 0.238 × 0.238 × 4.4

➝ Volume of cylinder = 22/7 × 0.056644 × 4.4

➝ Volume of cylinder = 22/7 × 0.2492336

➝ Volume of cylinder = 0.782 cm³

Henceforth, volume of cylinder = 0.782 cm³

${\large{\bold{\rm{\underline{Additional \; knowledge}}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: circle = \: \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Circumference \: of \: circle \: = \: 2 \pi r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto CSA \: of \: sphere \: = \: 2 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto SA \: of \: sphere \: = \: 4 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: sphere \: = \: 3 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Diameter \: of \: circle \: = \: 2r}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Radius \: of \: circle \: = \: \dfrac{d}{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: sphere \: = \: \dfrac{4}{3} \pi r^{3}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: rectangle \: = \: Length \times Breadth}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: rectangle \: = \:2(length+breadth)}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Perimeter \: of \: square \: = \: 4 \times sides}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Area \: of \: square \: = \: Side \times Side}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Volume \: of \: cylinder \: = \: \pi r^{2}h}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Surface \: area \: of \: cylinder \: = \: 2 \pi rh + 2 \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Lateral \: area \: of \: cylinder \: = \: 2 \pi rh}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Base \: area \: of \: cylinder \: = \: \pi r^{2}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Height \: of \: cylinder \: = \: \dfrac{v}{\pi r^{2}}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto Radius \: of \: cylinder \: = \:\sqrt frac{v}{\pi h}}}}$

$\; \; \; \; \; \; \;{\sf{\bold{\leadsto TSA \: of \: cuboid \: = \: 2(l \times b + b \times h + l \times h}}}$