Answer:

To find :-

Circumference of the circular swimming pool.

Given :-

Diameter of the circular swimming pool = 42 meters

How can we find the αnswer :-

Here we have diameter of the circular swimming pool firstly we will convert diameter of the circular swimming pool to radius of the circular swimming pool and the by using the formula .

Let's do

\boxed{\red{\:\mathsf{Radius \:=\:\dfrac{Diameter}{2}}}}

Radius=

2

Diameter

Radius = 42/2

Radius = 21 meters

Finding circumference of the circular swimming pool :-

\longmapsto\mathsf{Circumference \:of\: the\: circle\: =\: 2\:\pi \:r}⟼Circumferenceofthecircle=2πr

\longmapsto\mathsf{Circumference \:of\: the\: circle\: =\: 2\times\:\dfrac{22}{7}\:\times\:21}⟼Circumferenceofthecircle=2×

7

22

×21

\longmapsto\mathsf{Circumference \:of\: the\: circle\: =\: 2\times\:22\:\times\:3}⟼Circumferenceofthecircle=2×22×3

\longmapsto\mathsf{Circumference \:of\: the\: circle\: =\: 44\:\times\:3}⟼Circumferenceofthecircle=44×3

\longmapsto\mathsf{Circumference \:of\: the\: circle\: =\:132m }⟼Circumferenceofthecircle=132m

So the circumference of the circular swimming pool is 132m

αnswer :-

\boxed{\mathsf{\bigstar\:\:Circumference\:of\:the\:circular\:swimming\:pool\:=\:132m\:\:\bigstar}}

★Circumferenceofthecircularswimmingpool=132m★

\large{ \textsf{\textbf {\PINK {➸ \: \: QUESTION \: :-}}}}➸ Question :-

50 identical circular plates are placed one above the other to form a solid cylinder. For each plate, diameter = 21cm and thickness = 1.5cm.

Find :-

Curved surface area of the cylinder formed.

Volume of the cylinder formed.

\begin{GATHERED} \\ \end{gathered}

\large {\textsf{ \textbf {\pink {➸ \: \: Solution \: :-}}}} ➸ Solution :-

Given :-

50 plates each with diameter = 21cm

Radius = \sf \cfrac {21}{2}

2

21

cm

Thicknesses of each plate = 1.5cm

Total thickness of the plates = 1.5cm × 50 = 75cm

\boxed {\RM \bold \blue {CSA \: of \: a \: cylinder=2\pi rh}}

CSAofacylinder=2πrh

\blue ➸ \: \: \: \sf 2( \cfrac{22}{7} )( \dfrac{21}{2} )(75)➸2(

7

22

)(

2

21

)(75)

\blue ➸ \: \: \: \sf\cancel{ 2} \times \cfrac{22}{7} \times \cfrac{21}{\cancel2} \times 75➸

2

×

7

22

×

2

21

×75

\blue ➸~~~\sf \cfrac{22 \times 21 \times 75}{7}➸

7

22×21×75

\blue ➸ \: \: \: \sf \cfrac{34,650}{7}➸

7

34,650

\blue ➸~~~\sf CSA = 4950~cm^2➸ CSA=4950 cm

2

\boxed {\bold \blue{Volume \: of \: a \: cylinder =\pi r^2h} }

Volumeofacylinder=πr

2

h

\blue➸ \: \: \: \sf \cfrac{22}{7} ( \cfrac{21}{2} )( \cfrac{21}{2} )(75)➸

7

22

(

2

21

)(

2

21

)(75)

\blue ➸ \: \: \: \sf \cfrac{\cancel{22}}{7} \times \cfrac{21}{\cancel2} \times \cfrac{21}{2} \times 75➸

7

22

×

2

21

×

2

21

×75

\blue➸ \: \: \: \sf \cfrac{3,63,825}{14}➸

14

3,63,825

\blue➸ \: \: \: \sf Volume=25987.5~cm^3➸Volume=25987.5 cm

3

\begin{gathered} \\ \end{gathered}

\large {\textsf {\textbf \pink {➸ \: \: Know \: more \: :-}}}➸ Know more :-

Volume of cube = \sf a^3a

3

Volume of cuboid = \sf lbhlbh

CSA of cube = \sf 4a^24a

2

CSA of cuboid = \sf 2h(l+b)2h(l+b)

LSA of cube = \sf 6a^26a

2

LSA of cuboid = \sf 2(lb + bh + hl)2(lb+bh+hl)

\begin{gathered} \\ \end{gathered}

_____________________

@Sita05✨