Question

a wire is in a circular shape of area 5544cm square if the same wire is bent in the shape wire is bent in the shape of a square find the length of a side of the square and its area ​

Answer 1

Step-by-step explanation:

Answer

The area of circular shape is

5544cm

2

πr

2

=5544

7

22

r

2

=5544

r

2

=5544×

22

7

r

2

=252×7

r

2

=1764

r=

1764

r=42

The circumference of circle

=2πr

=2×

7

22

×42

=2×22×7

=154cm

Now, The perimeter of square is

154cm

Let the side be a,

∴4a=154

a=

2

77

Area of square

a

2

=(

2

77

)

2

=

4

5929

Answer 2

Step-by-step explanation:

$\frak Given = \begin{cases} &\sf{The\ area\ of\ the\ circular\ shaped\ wire\ =\ 5544cm².} \\ &\sf{The\ same\ wire\ is\ bent\ in\ the\ same\ of\ square.} \end{cases}$

To find:- We have to find the length of a side of the square and its area ?

__________________

$\frak{\underline{\underline{\dag As\ we\ know\ that:-}}}$

$\sf : \implies {\underline{\boxed{\sf Area_{(circular\ wire)}\ =\ πr².}}}$

__________________

$\frak{\underline{\underline{\dag By\ substituting\ the\ values,\ we\ get:-}}}$

$\sf : \implies {πr²\ =\ 5544} \\ \\ \sf : \implies {\dfrac{22}{7}\ r²\ =\ 5544} \\ \\ \sf : \implies {r²\ =\ 5544\ ×\ \dfrac{7}{22}} \\ \\ \sf : \implies {r²\ =\ 252\ ×\ 7} \\ \\ \sf : \implies {r²\ =\ 1764} \\ \\ \sf : \implies {r\ =\ \sqrt{1764}} \\ \\ \sf : \implies {\purple{\underline{\boxed{\bf r\ =\ 42.}}}}\bigstar$

$\sf \therefore {\underline{The\ radius\ of\ the\ circular\ wire\ is\ 42.}}$

Now here:-

$\frak{\underline{\underline{\dag As\ we\ know\ that:-}}}$

$\sf : \implies {\underline{\boxed{\sf Circumference_{(circle)}\ =\ 2πr.}}}$

$\frak{\underline{\underline{\dag By\ substituting\ the\ values,\ we\ get:-}}}$

$\sf : \implies {2\ ×\ \dfrac{22}{7}\ ×\ 42} \\ \\ \sf : \implies {2\ ×\ 22\ ×\ 7} \\ \\ \sf : \implies {\purple{\underline{\boxed{\bf 154cm.}}}}\bigstar$

$\frak{\underline{\underline{\dag Now,\ the\ perimeter\ of\ square\ is\ 154cm.}}}$

☯️ Let the side be a.

$\sf \therefore {4a\ =\ 154} \\ \\ \sf : \implies {a\ =\ \dfrac{77}{2}}$

$\frak{\underline{\underline{\dag Now,\ area\ of\ square:-}}}$

$\sf : \implies {a²\ =\ \bigg (\dfrac{77}{2} \bigg)^2} \\ \\ \sf : \implies {\purple{\underline{\boxed{\bf \dfrac{5929}{7}}}}}\bigstar$

Hence verified ✔.