Question

Find the circumference of a circle whose radius is:-
a)10.5
b)3.5​

Answer 1

⛦Question :

To find the circumference of circle for the given radius.

⛦ Given :

• 1st Radius = 10.5
• 2nd Radius = 3.5

⛦ Solution :

a) Radius = 10.5

We know that,

★Circumference of circle = 2πr

•Putting the value,

$\sf{Circumference_{(circle)}=}\sf{2\times}\dfrac{22}{7}\times{10.5}$

$\sf{Circumference_{(circle)}=}\dfrac{44\times\cancel{105}^{ \ 21}}{7\times\cancel{10}^{ \ 2}}$

$\sf{Circumference_{(circle)}=}\dfrac{\cancel{44}^{ 22}\times\cancel{21}^3}{\cancel{7}^{ \ 1}\times\cancel2^{ \ 1}}$

$\sf{Circumference_{(circle)}=}\sf{22\times3}$

$\sf{Circumference_{(circle)}}\sf{=66}$

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b) Radius = 3.5

•Putting the value,

$\sf{Circumference_{(circle)}=}\sf{2\times}\dfrac{22}{7}\times{3.5}$

$\sf{Circumference_{(circle)}=}\dfrac{\cancel{44}^{22}\times\cancel{35}^5}{\cancel7^1\times\cancel{10}^5}$

$\sf{Circumference_{(circle)}=}\dfrac{22\times5}{5}$

$\sf{Circumference_{(circle)}=}\dfrac{\cancel{110}^{22}}{\cancel{5}^1}$

$\sf{Circumference_{(circle)}}\sf{=22}$

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Therefore,

Final Answer :

• a) = 66
• b) = 22

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More Information :

You must know that :

•Value of π(pie) = $\dfrac{22}{7}$

•Diameter of circle = 2r = 2 × r

•Area of circle = πr²

Answer 2

Step-by-step explanation:

$\frak Given = \begin{cases} &\sf{The\ circumference\ of\ a\ circle\ whose\ radius\ is:-} \\ &\sf{a).\ 10.5} \\ &\sf{b). 3.5} \end{cases}$

To find:- We have to find the circumference of a circle ?

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$\frak{\underline{\underline{\dag As\ we\ know\ that:-}}}$

$\sf : \implies {\pink{\underline{Circumference_{(circle)}\ =\ 2πr.}}}$

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a). 10.5

$\frak{\underline{\underline{\dag By\ substituting\ the\ values,\ we\ get:-}}}$

$\sf : \implies {2\ ×\ \dfrac{22}{7}\ ×\ 10.5} \\ \\ \sf : \implies {\dfrac{44\ ×\ 105}{7\ ×\ 10}} \\ \\ \sf : \implies {\dfrac{44\ ×\ 21}{7\ ×\ 2}} \\ \\ \sf : \implies {22\ ×\ 3} \\ \\ \sf : \implies {\purple{\underline{\boxed{\bf Circumference\ =\ 66.}}}}\bigstar$

$\sf \therefore {\underline{Hence,\ the\ required\ circumference\ is\ 66.}}$

b). 3.5

$\frak{\underline{\underline{\dag By\ substituting\ the\ values,\ we\ get:-}}}$

$\sf : \implies {2\ ×\ \dfrac{22}{7}\ ×\ 3.5} \\ \\ \sf : \implies {\dfrac{44\ ×\ 35}{7\ ×\ 10}} \\ \\ \sf : \implies {\dfrac{22\ ×\ 5}{5}} \\ \\ \sf : \implies {\cancel \dfrac{110}{5}} \\ \\ \sf : \implies {\purple{\underline{\boxed{\bf Circumference\ =\ 22.}}}}\bigstar$

$\sf \therefore {\underline{Hence,\ the\ required\ circumference\ is\ 22.}}$

$\frak{\underline{\underline{\dag Hence,\ the\ required\ circumferences\ are:-}}}$

$\sf : \implies {a).\ 66.}$

$\sf : \implies {b).\ 22.}$