Question

the area of a circle is 55.44m2
find its radius​

Answer 1

Answer:

Given:-

The area of circle is 55.44 m², find its radius.

To Find:-

The radius of a circle

Note:-

●》Area of circle = πr² ( here "r" is radius ) π = 22/7

●》When we rearrange a term from one side to another~

•> + becomes -

•> - becomes +

•> × becomes ÷

•> ÷ becomes ×

•> ² " square " becomes √

_______________________

●》Any square form will be also be in a single unit when it is in √ form , for example = √m² changes to 'm'

Solution:-

Area of circle = 55.44 m², radius = ? , let radius be "x"

☆ According to note first and second point~

▪︎$Area \ \ of \ \ circle = πr²$

▪︎$55.44m² = 22/7 × x²$

☆ For doing calculations, it needs to be rearranged from its side, accordingly to note third point~

▪︎$55.44m² × 7 = 22 × x²$

▪︎$388.08 m² = 22x²$

▪︎$388.08 m² ÷ 22 = x²$

▪︎$17.64 m² = x²$

☆ Square when rearranged it becomes √ and from the value, the square will be removed

▪︎$√17.64 m² = x$

☆ After doing calculations and accordingly to note third point~

▪︎$4.2 m = x$

Checking:-

Remeber - For checking the area of circle both side term should be equal that L.H.S = R.H.S and for checking we would apply the "x" or radius value.

▪︎$Area \ \ of \ \ circle = πr²$

▪︎$55.44 m² = 22/7 × 4.2² \ \ ?$

☆ After doing half calculation~

▪︎$55.44 m² = 388.08/7 \ \ ?$

☆ After doing whole calculations~

▪︎$55.44 m² = 55.44 m² \ \ ?$ ( by calculator, it is done. )

yes !

Hence,Radius = x =》4.2m

Answer:-

Hence, the radius of a circle = $4.2 m$

Answer 2

Answer:

$\large{\sf{\pmb{\underline{\purple{Given:-}}}}}$

• ● The area of a circle is 55.44 m²

$\large{\sf{\pmb{\underline{\purple{To \: Find :-}}}}}$

• ● Radius of Circle

$\large{\sf{\pmb{\underline{\purple{Formula \: Used:-}}}}}$

$\bigstar\underline{\boxed{\sf{\pink{Area \: of \: Circle ={\pi}{r}^{2}}}}}$

$\large{\sf{\pmb{\underline{\purple{Solution:-}}}}}$

${\underline{\frak{\pmb{\bigstar \: Here}}}}$

• ● Area of Circle = 55.44 m²
• ● Radius of Circle

${\underline{\frak{\pmb{\bigstar \: According \: To \: The \: Question}}}}$

${ : \implies{\sf{Area \: of \: Circle ={\pi}{r}^{2}}}}$

• ● Substituting the values

${ : \implies{\sf{55.44 \: {m}^{2} ={\pi}{r}^{2}}}}$

${ : \implies{\sf{55.44 \: {m}^{2} ={\dfrac{22}{7} } \times {r}^{2}}}}$

${ : \implies{\sf{\dfrac{55.44 \times 7}{22} = {r}^{2}}}}$

${ : \implies{\sf{\dfrac{\cancel{55.44} \times 7}{\cancel{22}} = {r}^{2}}}}$

${ : \implies{\sf{2.52 \times 7}= {r}^{2}}}$

${ : \implies{\sf{17.64 = {r}^{2}}}}$

${ : \implies{\sf{ \sqrt{17.64} = {r}}}}$

${ : \implies{\sf{r = 4.2 \: m}}}$

${\bigstar{\underline{\boxed{\sf{\pink{r = 4.2 \: m}}}}}}$

● Henceforth,The Radius of Circle is 4.2 m.

$\large{\sf{\pmb{\underline{\purple{Verification:-}}}}}$

${ : \implies{\sf{Area \: of \: Circle ={\pi}{r}^{2}}}}$

• ● Substituting the values

${ : \implies{\sf{55.44 \: {m}^{2} ={\pi}{r}^{2}}}}$

${ : \implies{\sf{55.44 \: {m}^{2} ={\dfrac{22}{7} } \times {(4.2)}^{2}}}}$

${ : \implies{\sf{55.44 \: {m}^{2} ={\dfrac{22}{7} } \times {(4.2 \times4.2) {m}^{2} }}}}$

${ : \implies{\sf{55.44 \: {m}^{2} ={\dfrac{22}{7} } \times {17.64 \: {m}^{2} }}}}$

${ : \implies{\sf{55.44 \: {m}^{2} ={\dfrac{22}{\cancel{7}}}\times{\cancel{17.64}\: {m}^{2} }}}}$

${ : \implies{\sf{55.44 \: {m}^{2} ={22}\times 2.52 \: {m}^{2} }}}$

${ : \implies{\sf{55.44 \: {m}^{2} =55.44 \: {m}^{2}}}}$

${\bigstar{\underline{\boxed{\sf{\pink{LHS=RHS }}}}}}$

● Hence Verified ✔

$\large{\sf{\pmb{\underline{\purple{Know \: More:-}}}}}$

$\begin{gathered}\small\begin{gathered}\bigstar \: \bf\underline{More \: Useful \: Formulae } \: \bigstar  \\ \begin{gathered}{\boxed{\begin{array} {cccc}{\sf{{\leadsto TSA \: of \: cube \: = \: 6(side)^{2}}}} \\  \\{\sf{{\leadsto LSA \: of \: cube \:= \: 4(side)^{2}}}}  \\  \\{\sf{{\leadsto Volume \: of \: cube \: = \: (side)^{3}}}} \\  \\ {\sf{{\leadsto Diagonal \: of \: cube \: = \: \sqrt(l^{2} + b^{2} + h^{2}}}} \\  \\ {\sf{{\leadsto Perimeter \: of \: cube \: = \: 4(l+b+h)}}} \\   \\ {\sf{{\leadsto CSA \: of \: sphere \: = \: 2 \pi r^{2}}}} \\  \\ {\sf{{\leadsto SA \: of \: sphere \: = \: 4 \pi r^{2}}}} \\  \\{\sf{{\leadsto TSA \: of \: sphere \: = \: 3 \pi r^{2}}}} \\  \\ {\sf{{\leadsto Diameter \: of \: circle \: = \: 2r}}} \\  \\ {\sf{{\leadsto Radius \: of \: circle \: = \: \dfrac{d}{2}}}} \\  \\ {\sf{{\leadsto Volume \: of \: sphere \: = \: \dfrac{4}{3} \pi r^{3}}}} \\  \\ {\sf{{\leadsto Area \: of \: rectangle \: = \: Length \times Breadth}}} \\  \\ {\sf{{\leadsto Perimeter \: of \: rectangle \: = \:2(length+breadth)}}} \\  \\{\sf{{\leadsto Perimeter \: of \: square \: = \: 4 \times sides}}}\end{array}}}\end{gathered}\end{gathered}\end{gathered}$