Question

6. A wire is in the shape of a circle of radius 21 cm. It is bent in the form of a square. The side of the square is { use π = 22/7 }
(a) 22 cm
(b) 44 cm
(c) 66 cm
(d) 33 cm

Explain your answer
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Answer 1

Answer:

Option (d) 33 cm is the correct answer.

Step-by-step explanation:

As, length of wire is same so, both figures have same perimeter.

Therefore,

Circumference of circle = Perimeter of the square.

Let r be the radius of the circle and a is the side of the square.

So, 2πr = 4a

• 4a = 2 × 22/7 × 21 ( r = 21 cm )
• 4a = 132
• a = 33 cm

hope it helps you.

Answer 2

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

Given that,

A wire is in the shape of a circle of radius 21 cm. It is bent in the form of a square.

Since, wire is in the shape of a circle of radius 21 cm.

Let r represents the radius of circle.

So, Radius of circle, r = 21 cm

We know,

Circumference of a circle of radius r is given by

$\boxed{\sf{ \: \: Circumference_{(circle)} \: = \: 2 \: \pi \: r \: \: }}$

So, on substituting the value of r, we get

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

$\rm \: Circumference_{(circle)} \: = \: 2 \times 22 \times 3$

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

Let assume that side of square be x units.

Now, the circular wire is bent in the form of square.

So, it implies Perimeter of square is equals to Circumference of circle.

$\rm \: Perimeter_{(square)} \: = \: Circumference_{(circle)}$

$\rm \: 4x = 132$

$\rm \: x = \dfrac{132}{4}$

$\rm\implies \:x \: = \: 33 \: cm$

Hence,

Side of square, x = 33 cm

So, Option (d) is correct.

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