Question

The width of a rectangular field is 2/3 of its length. Find its dimensions if the perimeter of the field is 220m.​

Answer 1

Given :-

• The width of a rectangular field is 2/3 of its length.

• The perimeter of the field is 220m.

To find :-

• Dimensions of rectangular field

Solution :-

Let the length be x then its width be 2x/3

• Perimeter of rectangular field = 220m

As we know that

Perimeter of rectangle = 2(l + b)

Where " l " is length and " b " is breadth of rectangular field.

Now as per given condition

→ Perimeter of rectangular field = 220

→ 2(l + b) = 220

→ 2(x + 2x/3) = 220

→ x + 2x/3 = 220/2

→ 3x + 2x/3 = 110

→ 5x/3 = 110

→ 5x = 3 × 110

→ 5x = 330

→ x = 330/5

→ x = 66

Hence,

• Dimensions of rectangular field

• Length = x = 66m

• Breadth = 2x/3 = 2 × 66/3 = 2 × 22 = 44m

Answer 2

AnswEr :-

• Length of the rectangular field is 66m.
• Breadth of the rectangular field is 44m.

Given :-

• The width of a rectangular field is 2/3 of its length. The perimeter of the field is 220m.

To Find :-

• Dimensions of the field.

SoluTion :-

Let,

• Length of the rectangular field be x
• Breadth of the rectangular field will be 2x/3

Perimeter of the field is 220m

As we know that the perimeter of the rectangle is

2 (l + b)

Where,

• l = length
• b = breadth

According to question :-

2 (x + 2x/3) = 220

→ x + 2x/3 = 220/2

→ (3x + 2x) /3 = 110

→ 5x/3 = 110

→ 5x = 110 × 3

→ 5x = 330

→ x = 330/5

→ x = 66

• Length of the rectangular field => x => 66m
• Breadth of the rectangular field => 2x/3 => 2 × 22 = 44m

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