[Majida]

Given:
Width of the rectangle = 5 units
Perimeter of the rectangle = 120 units

Let’s assume the length of the rectangle is “x” times its width.

The formula for the perimeter of a rectangle is given by:
Perimeter = 2 × (Length + Width)

Putting in the given values, we have:
120 = 2 × (x * 5 + 5)

Simplifying, we get:
120 = 2 × (5x + 5)
120 = 10x + 10
10x = 120 – 10
10x = 110
x = 110/10
x = 11

Therefore, the length of the rectangle is 11 times its width.

Now, let’s calculate the length:
Length = x * Width Length = 11 * 5 Length = 55 units

To find the area of the rectangle, we use the formula:
Area = Length × Width

Putting in the values, we get:
Area = 55 × 5 Area = 275 square units

Hence, the area of the rectangle is 275 square units.

[Majida]

Let the breadth of the rectangle be “b”.

The length of the rectangle is twice the breadth, so it’s “2b”.

The perimeter of the rectangle is 2x(b+2b) =6b.

The wire bent in the form of a circle also has a circumference of 6b.

Let “r” be the radius of the circle, that is,

2πr=6b, implies, r= 6b/2π

The area of the rectangle is 2b², and the area of the circle is 9b².

So, according to the question, we have,

Area of the circle – Area of the rectangle = 104.5

9b² – 2b² = 104.5

7b² = 104.5

b² = 104.5 / 7

b² = 14.93

b ≈ √14.93 = 3.86

The total length of the wire used is equal to the perimeter of the rectangle:

Total length of wire = 6b
= 6 × 3.86
= 23.16

Therefore, the length of the wire is approximately 23.16 cm.