A wire when bent in the form of a equilateral triangle encloses an area of 36 √3 sq.cm.Find the area enclosed by the same wire when bent to form a square and a rectangle whose length is 2 cm more than its width.

The correct answer is: ∴ Areas of the given Square and rectangle are 81 cm2 & 80 cm2, respectively.

Area of an equilateral triangle = (√3 / 4) × side²
Area of a square = Side²
Area of a rectangle = length × breadth
Step-by-step solution:-
Area of the triangle = (√3 / 4) × side²
∴   36 √3 = (√3 / 4) × side² …................................................ (From given information)
∴   36 = 1 / 4 × side²
∴   36 × 4 = side²
∴   144 = side²
∴   12 = side ................................................................................. (Taking square root both the sides)
i.e. Side of the triangle = 12 cm
Length of the wire used to make the triangle = Perimeter of this triangle = 3 × side = 3 × 12 = 36 cm ..................... (Equation i)
We Know that the same wire is used to make the square and the triangle.
∴ Perimeter of the square = perimeter of the triangle
∴ Perimeter of the square = 36 ...................................................................................................................... (From Equation i)
∴   4 × side = 36 .................................................................................. (Perimeter of a square = 4 × side)
∴   side = 36/4 = 9 cm ..................................................................................................... (Equation ii)
∴   Area of the square = side²
∴   Area of the square = 9² ........................................................................................................................ (From Equation ii)
∴   Area of the square = 81 cm²
For the rectangle,
Let the width be x cm
∴  Width = x cm ............................................................................................................................................ (Equation iii)
∴  Length = width + 2 cm = x + 2 ..................................................................................................................... (Equation iv)
We Know that the same wire is used to make the rectangle and the triangle.
∴  Perimeter of the rectangle = perimeter of the triangle
∴  Perimeter of the rectangle = 36 ...................................................................................................................... (From Equation i)
∴  2 (length + breadth) = 36 .................................................................................. (Perimeter of a rectangle= 2 × length + breadth)
∴  2 (x + x + 2) = 36 ..................................................................................................... (From Equations iii & iv)
∴  2 (2x + 2 ) = 36
∴  4x + 4 = 36
∴  4x = 36 - 4
∴  4x = 32
∴   x = 32/4 = 8 cm ................................................................................... (Equation v)
Substituting Equation v in Equations iii & iv, we get-
Width = x = 8
Length = x + 2 = 8 + 2 = 10
∴ Area of the rectangle = length × breadth
∴ Area of the rectangle = 8 × 10
∴ Area of the rectangle = 80 cm²
Final Answer:-
∴ Areas of the given Square and rectangle are 81 cm² & 80 cm², respectively.