Question

the wire bent in the form of equilateral triangle encloses a area of 36√3 find thae area of rectangle whose length is 2cm more than its width​

Answer 1

Answer:

$\frac{ \sqrt{3} }{4} {a}^{2} is \: the \: area \: of \: an \: equilateral \: triangle$

Step-by-step explanation:

Putting the formula,

the side of the ∆ is 12 cm.

Perimeter= 12×3 = 36cm.

the length is 2cm > the width of the rectangle.

let the width = x cm and length= (x+2)cm.

ATP,

2(length + width) = 36

or, 2(x + x + 2) = 36

or, x = 8.

length= 8+2= 10cm .

width= 8 cm.

area= 10×8cm^2 = 80cm^2