Question

a wire is bent to form a square of side 11cm if this wire is re should to form a rectangle of length 16 cm find the breadth of rectangle and area of rectangle
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Answer 1

Required Answer:-

A simple logic we have to apply here. We know that the same wire was used to make both the shapes. It means the length of the wire is same for both square and rectangle. Hence the perimeter of square = perimeter of rectangle$.$

Now,

We already have the side of the square, So we can find the perimeter of square because:

• Perimeter of square = 4 × Side.

Using this,

➛ Perimeter of square = 4 × 11 cm

➛ Perimeter of square = 44 cm

The perimeter of rectangle is also 44 cm as per above discussion. And we know that,

• Perimeter of rectangle = 2(Length + Breadth)

Then,

➛ 2(16 cm + Breadth) = 44 cm

➛ 16 cm + Breadth = 22 cm

➛ Breadth = 6 cm

And,

We have to find the area of the rectangle also.

• Area of rectangle = Length × Breadth

Using this,

➛ Area = 16 cm × 6 cm

➛ Area = 96 cm²

Hence:-

• Breadth of the rectangle = 6 cm
• Area of the rectangle = 96 cm²

And we are done! :D

Answer 2

Answer:

Given

• Side of square wire = 11 cm
• Length of wire = 16 cm

To Find :-

Breadth and area of rectangle

Answer :-

For finding perimeter we will first find perimeter of square.

$\sf \: Perimeter \: = 4 \times s$

$\sf \: Perimeter \: = 4 \times 11$

$\sf \: Perimeter = 44 \: cm$

Therefore :-

Perimeter of square wire = 44 cm

Now,

Perimeter of square = Perimeter of rectangle

Lets find breadth of rectangle

$\sf \: Perimeter \: of \: rectangle = 2(l + b)$

$\sf \: 44 = 2(16 + b)$

$\sf \: 44 = 32 + 2b$

$\sf \: 44 - 32 = 2b$

$\sf \: 12 = 2b$

$\sf \: b = 6 \: cm$

Hence :-

Breadth of rectangular wire is 6 cm

Now,

Let's find Area

$\huge \tt Area = l \times b$

$\sf \: Area = 16 \times 6$

$\sf \: Area = 96 \: cm {}^{2}$