The length and breadth of a rectangle are 10 cm and 6 cm respectively. The side of a square is 8 cm, compare the perimeters of the given rectangle and square and compare the areas of the given rectangle and square.

Given :

Length of a rectangle = 10 cm

Breadth of a rectangle = 6 cm

The length of the side of a square = 8 cm

To find :

We have to find the perimeter and area of the rectangle and square.

Solution :

We know that,

Perimeter of a rectangle of length l and breadth b is $2(l+b)$.

Area of a rectangle of length l and breadth b is lb.

Perimeter of a square of side s is 4s.

Area of a square of side s is s².

Therefore,

Perimeter of the given rectangle = $2(10+6) cm = 2(16) cm = 32 cm.$

Area of the given rectangle =$10\times6 cm^2 = 60 cm^2.$

Perimeter of the given square = $4(8) cm = 32 cm.$

Area of the given square = $8^2 cm^2 = 64 cm^2.$

Perimeter of the square and rectangle are equal.

Area of the square is greater than the area of the rectangle by $64 cm^2-60 cm^2 = 4 cm^2.$

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Updated on: 10-Oct-2022

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