a wire when bent in the form of an equilateral triangle encloses an area of 49 root 3 cm2, find the largest area enclosed by the same wire when bent to form : (i) a square (ii) a rectangle of length 12 cm.

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Please find below the solution to the asked query:

Given :a wire when bent in the form of an equilateral triangle encloses an area of 49 $\sqrt{3}$ cm² ,
We know , Area of equilateral triangle  = $\frac{\sqrt{3}}{4}\times {\left(\mathrm{Side}\right)}^2$ , So 
$\frac{\sqrt{3}}{4}\times {\left(\mathrm{Side}\right)}^2$ = 49 $\sqrt{3}$

Side ² =  196 

Side ² =  14²

Side  = 14

Here

Length of wire  =  Perimeter of given equilateral triangle  , So

Length of wire  = 14 + 14 + 14  =  42 cm

i ) When that wire bent into a square then length of wire is equal to perimeter of square .

We know perimeter of square  =  4 ( Side )  , So

4 ( Side ) =  42

Side = 10.5

And we know area of square = ( Side )²

So,

Area of that square  = ( 10.5 ) ² = 110.25 cm²                                           ( Ans )

ii ) When that wire bent into a rectangle then length of wire is equal to perimeter of rectangle .

We know perimeter of rectangle =  2 ( Length + Width )  , So

2 ( Length + Width ) =  42

Length + Width = 21 , given length  =  12 cm , so


12 + Width  =  21

Width  =  9 cm

And we know area of rectangle  = Length $\times$ Width


So,

Area of that rectangle = 12 $\times$ 9 = 108 cm²                                           ( Ans )


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