Question

shazli took a wire of length 44cm and bent it into shape of a circle.also find it's area . if the same wire is bent into shape of a square , what will be the length of each sides ? which figure encloses more area circle or square ?(take π=22/7)​

Answer 1

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Answer 2

$\sf\underbrace{Appropriate\:Question: }$

${: \longrightarrow}$shazli took a wire of length 44cm and bent it into shape of a circle.also find it's area . if the same wire is bent into shape of a square , what will be the length of each sides ? which figure encloses more area circle or square ?(take π=22/7)

$\sf\underbrace{understanding~the~question }$

• $\huge\mathfrak\orange\star$In the question, we are provided with the length of wire, and we need to find the radius + area of a circle and also if the same wire is bent into square, then we have to find it's length. So first, we will take radius as r, then we will apply the formula of circumference of circle to find the radius, after that we will apply the formula of area of circle to find the area of the circle. After that we will find the length of each side of the square, by applying the formula of perimeter of square.

$\bf\underline{let's~do}$

Given:-

• Length of Wire = 44cm

To Find:-

• Radius of the Circle
• Area of the Circle
• Perimeter of the Square
• Area of the Square

Which encloses more area ( ○ or □ )

Using Formulas:-

Circumference of Circle = ${2 \pi r}$

Area of Circle = ${ \pi r²}$

Perimeter of Square = 4 × Side

Area of Square = Side × Side

Solution with step by Step Explanation:-

Here's Questions Say that take $\pi$ as $\sf\dfrac{22}{7}$

$\bf\red{so,}$

Circumference of Circle = 44

$\begin{gathered}\sf:\implies{ 2 \pi r = 44cm}\\\\\\ :\implies\sf{2 \times \sf\dfrac{22}{7} \times r = 44 }\\\\\\ :\implies\sf{r = {\cancel {44}^{2}} \times \dfrac{1}{\cancel{2}} \times \dfrac{7}{\cancel{22_1}} = 7cm } \\ \\ \: {\underline{\sf{\orange{~Now, we ~have~to~find~the~area~of~Circle}}}}\\\\\\:\implies\sf{Area~of~Circle = \pi r²}\\\\\\:\implies\sf{ \dfrac{22}{\cancel{7}}\times {\cancel{7}}\times 7= 154cm²}\\\\\\ \end{gathered}$

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$\begin{gathered}{\underline{\sf{\orange{Now, Lets~find~perimeter ~of ~Square}}}}\\\\\\:\implies\sf{Perimeter ~of~Square = 44} \\ \\ :\implies \sf{4\times Side = 44}\\\\:\implies \sf{Side = \dfrac{44}{4} = 11cm}\\\\ \dag\: {\underline{\sf{\pink{Now~let's ~find~the ~area~of~circle}}}}\\\\\\ :\implies \sf{Area~of~Square = Side \times Side }\\\\:\implies\sf{11 \times 11 }\\\\:\implies\sf{121cm²}\\\\\\\\\end{gathered}$

${\underline {\frak{~~~154cm² > 121 cm²}}}$

$\therefore {\underline{\rm{\green{Circle~ encloses ~more~ Area ~ than~ square }}}}$

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