Question

A cylindrical well has a concrete wall,14 cm thick, and its depth is 20 cm. If the diameter of the outer circle of the wall is 3.08 m. Find the area of its inner curved surface

Answer 1

$\: \: \: \:{\large{\bold{\sf{\underbrace{Understanding \; the \; question}}}}}$

➨ This question says that there is a cylindrical well and it has a concrete wall. And it is 14 cm thick and 20 cm deap. Actually in short thickness is 14 cm and height is 20 cm ( atq ). Now this question says that the diameter of outer circle of that wall is about 3.08 metres. Now we have asked to find the area of its inner curved surface ! ( Yeah..! nice question let's find our answer )

$\: \: \: \:{\large{\bold{\sf{\underline{Given \; that}}}}}$

➨ Thickness = 14 cm

➨ Height or depth = 20 cm

➨ Diameter of outer circle of that wall = 3.08 metres

$\: \: \: \:{\large{\bold{\sf{\underline{To \; find}}}}}$

➨ Area of inner curved surface of that wall of the well.

$\: \: \: \:{\large{\bold{\sf{\underline{Solution}}}}}$

➨ Area of inner curved surface of that wall of the well = 1.232 m³

$\: \: \: \:{\large{\bold{\sf{\underline{Using \; concept}}}}}$

➨ Formula to find Curved Surface Area

➨ Formula to find radius

$\: \: \: \:{\large{\bold{\sf{\underline{Using \; formula}}}}}$

➨ Curved Surface Area = πr²h

➨ Radius = ${\bold{\sf{\dfrac{Diameter}{2}}}}$

$\: \: \: \:{\large{\bold{\sf{\underline{These \; denotes \; or \; means}}}}}$

➨ r denotes radius

➨ h denotes height

➨ π pronounced as pi

➨ The value of π is ${\bold{\sf{\dfrac{22}{7}}}}$

➨ Curved Surface Area means CSA

➨ Height means h

$\: \: \: \:{\large{\bold{\sf{\underline{Full \; Solution}}}}}$

_________________

↦ Finding radius of outer circle of that wall.

~ Given diameter is 3.08 metres

➥ Radius = ${\bold{\sf{\dfrac{Diameter}{2}}}}$

➥ Radius = ${\bold{\sf{\dfrac{3.08}{2}}}}$

➥ Radius = ${\bold{\sf{1.54 \: m}}}$

↦ Now let's convert metres into centimetres

~ 1.54 metres (given)

➥ 1 m = 100 cm

➥ 1.54 × 100

➥ 154 cm

• Henceforth, the outer radius of circle of the wall is 154 cm.

↦ Finding inner radius

➥ Outer radius - Thickness

➥ 154 - 14

➥ 140 cm

Henceforth, inner radius is 140 cm

________________

I know guys you are little bit confused due to why I change diameter into radius and metres into centimetres.

No need to worry let me tell you !

↦ I change diameter into radius because our using formula says that Curved Surface Area = πr²h and here there is no donation for diameter but it's here for radius that's why there is a need to convert diameter into radius !

↦ I change metres into centimetres because of a simple reason and it's that we have to work according to the question always. And the question have units ( m and cm ) both that's why there is a need to change one of them !

But there is twist in this question that the diameter of the outer circle of the wall is given in metres but we change the radius in cm so at last we have to change all the unit of cm in m again' !

________________

↦ Converting all the cm units in m

~ Height = 20 cm

➥ 1 cm = 1/100 m

➥ 20/100

➥ 0.2 metres

~ Inner radius = 140 cm

➥ 1 cm = 1/100 m

➥ 140/100

➥ 1.4 metres

____________________

↦ Now at last let's find the area of its inner curved surface

➥ CSA = πr²h

➥ CSA = ${\bold{\sf{\dfrac{22}{7}}}}$ × 1.4² × 0.2

➥ CSA = ${\bold{\sf{\dfrac{22}{7}}}}$ × 1.4 × 1.4 × 0.2

• Cancelling 1.4 by 7 we get 0.2

➥ CSA = 22 × 0.2 × 1.4 × 0.2

➥ CSA = 22 × 0.2 × 0.2 × 1.4

• ( Continuing......... )

➥ CSA = 1.232 m³

• Henceforth, 1.232 m³ is the area of its inner curved surface.

Answer 2

Answer :-

• Inner CSA of the Well = 1.232 m²

Given :-

• Thickness of Well = 14 cm = 0.14 m
• Depth (Height) of Well = 20 cm = 0.20 m
• Outer Diameter of Well = 3.08 m

To Find :-

• Inner CSA of the Well = ?

Explanation :-

In order to find the Inner CSA of well, First we need to Find the Outer Radius of Well.

Finding the Outer Radius of Well,

$\sf \implies Outer \: Radius = \dfrac{Diameter}{2}$

$\sf \implies Outer \: Radius = \dfrac{3.08}{2}$

$\blue { \sf \implies Outer \: Radius = \bold{1.54 \: m} }$

Now we have Outer Radius of well that is 1.54 m. Now, we need to find the Inner Radius of Well to find the inner CSA of well.

Finding the Inner Radius of Well,

$\sf \implies Inner\: Radius = Outer \: Radius - Thickness$

$\sf \implies Inner\: Radius = 1.54 - 0.14$

$\purple { \sf \implies Inner\: Radius = \bold{1.4 \: m} }$

Now Finding the CSA of Inner Well,

$\green { \boxed { \bf \bigstar \: CSA \: of \: Inner \: well = \pi r^2h \: \: cm^2 \: \bigstar }}$

$\sf : \: \longmapsto \pi r^2h = Inner \: CSA \: of \: Well$

$\sf : \: \longmapsto \dfrac{22}{7} \times (1.4)^2 \times 0.20$

$\sf : \: \longmapsto \dfrac{22}{7} \times 1.4 \times 1.4 \times 0.20$

$\sf : \: \longmapsto 22 \times 0.2 \times 1.4 \times 0.20$

$\sf : \: \longmapsto 4.4 \times 0.28$

$\red { \underline { \boxed { \sf : \: \longmapsto 1.232 \: m^2 }}}$

Hence, the Inner CSA of the Well is 1.232 m².