Sumit took a wire of length 44 cm and bent it into the shape of a circle. If the same wire is bent into the shape of a square, then find the length of the side of the square and also the figure which encloses more area. (Take π $=\frac{22}{7}$)

Given:

Length of the wire = $\mathbf{44}\mathbf{ }\mathbf{cm}$

When the wire is bent into the shape of a circle,

Let the radius be $\mathbf{r}\mathbf{ }\mathbf{cm}\mathbf{.}$

So, the circumference of the circle = length of the wire

$\Rightarrow 2\mathrm{\pi r} = 44$ 

$\Rightarrow \mathrm{\pi r} = \frac{44}{2}$ 

$\Rightarrow \mathrm{r} = \frac{22}{\mathrm{\pi }}$ 

$\Rightarrow \mathrm{r} = \frac{22\times 7}{22}$ 

$\Rightarrow \mathbf{r}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{7}$

So,$\mathbf{ }\mathbf{radius}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{7}\mathbf{ }\mathbf{cm}$

Now,

We know that the area of the circle = ${\mathbf{\pi r}}^{\mathbf{2}}$

                                  = $\frac{22}{7}\times 7\times 7$  = $\mathbf{154}\mathbf{ }{\mathbf{cm}}^{\mathbf{2}}$

So, the area of the circle is $\mathbf{154}\mathbf{ }{\mathbf{cm}}^{\mathbf{2}}$.

Now,

When the wire is bent into the shape of a square

Therefore, the perimeter of the square = length of the wire

$\Rightarrow 4\times \mathrm{Side} = 44$ 

$\Rightarrow \mathrm{Side} = \frac{44}{4}$

$\mathbf{\Rightarrow }\mathbf{Side}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{11}\mathbf{ }\mathbf{cm}$

Now, the area of square = Side²

                                            = (11)² = $\mathbf{121}\mathbf{ }{\mathbf{cm}}^{\mathbf{2}}$

Hence, the circle encloses more area by (154 - 121) cm² = $\mathbf{33}\mathbf{ }{\mathbf{cm}}^{\mathbf{2}}\mathbf{.}$