Question

A wire in the form of a square encloses an area
of 144 cm2 How much area is enclosed if the
same wire is bent in the form of a rectangle of
length 16cm

Answer 1

Answer:

The area enclosed is 128 cm².

Step-by-step explanation:

Let length of the side be ‘a’.

➠ a² = 144

➠ a = √144

➠ a = 12cm

Which implies perimeter of the square is 4 × 12 = 48cm.

So, the perimeter of the rectangle is 48cm = 2(l + b)

➠ 2(16 + b)

➠ 48 = 2(16 + b)

➠ 24 = 16 + b

➠ b = 8cm

∴ Therefore, Breadth of the rectangle is 8cm.

Now,

Area of a rectangle = Length × Breadth.

➠ 16 × 8

➠ 128 cm²

∴ Area of the rectangle enclosed is 128 cm².

Answer 2

Given

• Area of square = 144 cm²
• Square (wire) bent to form rectangle of,
• Length = 16 cm

To find

• Area enclosed when bent

Solution

• s² = 144
• s = √144
• s = 12 cm

Hence, perimeter of square = 4(s) = 4(12) = 48 cm

According to the question :-

→ Perimeter of square = Perimeter of rectangle

• 48 = 2(L + B)
• 48 = 2(16 + B)
• 48 = 32 + 2B
• B = 48 - 32
• 2B = 16
• B = 16/2
• B = 8

Hence, the breadth of rectangle is 8 cm

Now, area of rectangle enclosed,

• Area = Length × Breadth
• Area = 16 × 8
• Area = 128 cm²

Hence, the area is 128 cm²