Question

1. find the circumference of the circle whose radius is:(take
$\pi = \frac{22}{7}$
1. 14cm
2. 21dm
3. 35cm
4. 56dm​

Answer 1

$\large\textbf{\underline{\fbox{Given:}}}$

$\rm{Radius = \pi( \frac{22}{7}) } \\ \textrm{Circumference = 2} {\pi} \: \rm{r} \\ { = 2 \times (\frac{22}{7}) \times (\frac{22}{7})} \\ = \rm{ \frac{22 \times 7}{2 \times 22} } = \frac{2 \times 7}{2 \times 2} \\ = \frac{14}{4} = \frac{14}{ \: \: \: \cancel4 \: \: 1}$

$\large\textbf{\underline{\fbox{Answer:}}}$

Hence 4’s table doesn't have 14 in it we’re not gonna cancel it. Therefore the answer is 14cm.

$\huge{\fcolorbox{lavender}{lavender}{\purple{\#Be Brainly}}}$

Answer 2

$\large\underline{\sf{Solution-1}}$

Given that, radius of circle, r = 14 cm

We know,

Circumference of a circle of radius r is given by

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{Circumference_{(circle)}\:=\:2\:\pi\:r \: }}}}}} \\ \end{gathered}$

So, on substituting the value, we get

$\rm \: Circumference_{(circle)} \: = \: 2 \times \dfrac{22}{\cancel 7} \times \cancel {14} {} \: \: ^{2}$

$\rm\implies \:Circumference_{(circle)} \: = \: 88 \: cm$

$\large\underline{\sf{Solution-2}}$

Given that, radius of circle, r = 21 dm

We know,

Circumference of a circle of radius r is given by

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{Circumference_{(circle)}\:=\:2\:\pi\:r \: }}}}}} \\ \end{gathered}$

So, on substituting the value, we get

$\rm \: Circumference_{(circle)} \: = \: 2 \times \dfrac{22}{\cancel 7} \times \cancel {21} {} \: \: ^{3}$

$\rm\implies \:Circumference_{(circle)} \: = \: 132 \: dm$

$\large\underline{\sf{Solution-3}}$

Given that, radius of circle, r = 35 cm

We know,

Circumference of a circle of radius r is given by

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{Circumference_{(circle)}\:=\:2\:\pi\:r \: }}}}}} \\ \end{gathered}$

So, on substituting the value of r, we get

$\rm \: Circumference_{(circle)} \: = \: 2 \times \dfrac{22}{\cancel 7} \times \cancel {35} {} \: \: ^{5}$

$\rm\implies \:Circumference_{(circle)} \: = \: 220 \: cm$

$\large\underline{\sf{Solution-4}}$

Given that, radius of circle, r = 56 dm

We know,

Circumference of a circle of radius r is given by

$\begin{gathered}\red\bigstar\:\:{\underline{\orange{\boxed{\bf{\green{Circumference_{(circle)}\:=\:2\:\pi\:r \: }}}}}} \\ \end{gathered}$

So, on substituting the value of r, we get

$\rm \: Circumference_{(circle)} \: = \: 2 \times \dfrac{22}{\cancel 7} \times \cancel {56} {} \: \: ^{8}$

$\rm\implies \:Circumference_{(circle)} \: = \: 352 \: dm$

So, Circumference when radius is given are as follow

$\begin{gathered}\boxed{\begin{array}{c|c} \bf radius & \bf Circumference_{(circle)} \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 14 \: cm & \sf 88 \: cm \\ \\ \sf 21 \: dm & \sf 132 \: dm \\ \\ \sf 35 \: cm & \sf 220 \: cm\\ \\ \sf 56 \: dm & \sf 352 \: dm \end{array}} \\ \end{gathered}$

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Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$