According to question:-

✠ Given that:-

• Circumference of circle=Perimeter of square.

✠ To find:-

• Radius of the circle

✠ Formula used:-

[tex]\large \underline{\boxed{\sf{ \red{\implies2\pi r = circumference_{circle}}}}}[/tex]

Where,

• π = 22/7 or 3.14
• r=radius

Solution:-

[tex]\large\underline{\sf{ \implies2\pi r = 154 {cm}}}[/tex]

[tex]\large\underline{\sf{ \implies2 \times \frac{22}{7} \times r = 154cm }}[/tex]

[tex]\large\underline{\sf{ \implies \frac{44}{7} \times r = 154 {cm}}}[/tex]

[tex]\large\underline{\sf{ \implies r = \frac{154 {cm} \times 7}{44} }}[/tex]

[tex]\large\underline{ \boxed{\sf{ \red{\maltese\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: r = 24.5cm}}}}[/tex]

______________________________________

Additional information!!

[tex]\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \quad \footnotesize\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\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 =Breadth\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}

FormulasofAreas:−

⋆Square=(side)

2

⋆Rectangle=Length×Breadth

⋆Triangle=

2

1

×Breadth×Height

⋆Scalene△=

s(s−a)(s−b)(s−c)

⋆Rhombus=

2

1

×d

1

×d

2

⋆Rhombus=

2

1

d

4a

2

−d

2

⋆Parallelogram=Breadth×Height

⋆Trapezium=

2

1

(a+b)×Height

⋆EquilateralTriangle=

4

3

(side)

2

[/tex]

Verified answer

Answer:

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Explanation:

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