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# 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) × side^{2}

Area of a square = Side^{2}

Area of a rectangle = length × breadth

Step-by-step solution:-

Area of the triangle = (√3 / 4) × side^{2}

∴ 36 √3 = (√3 / 4) × side^{2} …................................................ (From given information)

∴ 36 = 1 / 4 × side^{2}

∴ 36 × 4 = side^{2}

∴ 144 = side^{2}

∴ 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 = side^{2}

∴ Area of the square = 9^{2} ........................................................................................................................ (From Equation ii)

∴ Area of the square = 81 cm^{2}

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 cm^{2}

Final Answer:-

∴ Areas of the given Square and rectangle are 81 cm^{2} & 80 cm^{2}, respectively.

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