Question

A wire is in the shape of a square of side 10cm. If the wire is re bent into a rectangle of length 12 cm, find its breadth. Which encloses more area?

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Answer 1

Answer:

Let a be the side of a square, l be the length and b be the breadth of rectangle.

As length of wire is constant, perimeter of both shapes would be same,

⇒4a = 2(l+b)

⇒40 = 2(12+b)

⇒b = 18 cm

Area of square = a² = 10² = 100cm²

Area of rectangle = lb = 12×118 = 2116cm²

⇒ Area of rectangle is greater by 116cm²

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Answer 2

Given

• A wire is in the shape of a square of side 10 cm
• The wire is rebent to form a rectangle.
• Length of rectangle is 12 cm

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To Find

• The breadth of rectangle.
• The shape which encloses more area.

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Solution

The perimeter of the wire bent into a square would be the total length of the wire.

So let's find the perimeter of the square.

Perimeter of Square → 4 × Side

Perimeter of Square → 4 × 10

Perimeter of Square → 40 cm

Perimeter of Square = Perimeter of Rectangle

Perimeter of Rectangle → 40 cm

Lenght → 12 cm

Breadth → x cm

Formula to find Perimeter of Rectangle → 2 (Lenght + Breadth)

Let's solve the below equation to find the breadth.

2 (12 + x) = 40

Step 1: Simplify the equation.

⇒ 2 (12 + x) = 40

⇒ 24 + 2x = 40

Step 2: Subtract 24 from both sides of the equation.

⇒ 24 + 2x - 24 = 40 - 24

⇒ 2x = 16

Step 3: Divide 2 from both sides of the equation.

⇒ 2x ÷ 2 = 16 ÷ 2

⇒ x = 8

∴ The breadth of the rectangle is 8 cm

Now let's find the area of the square and rectangle.

Area of Square → (Side)²

Area of Square → (10)²

Area of Square → 100 cm²

Area of Rectangle → Lenght × Breadth

Area of Rectangle → 12 × 8

Area of Rectangle → 96 cm²

∴ 100 > 96

100 - 96 = 4

∴ The square encloses more area by 4 cm² than the rectangle.

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