Question

A wire is in the shape of a rectangle its length is 40 cm and breadth is 22 CM if the same wire is rebent in the shape of a square what will be the measures of each side also find which shape encloses more area and by how much

Answer 1

HEYA,

Here is your answer

L=40cm
B=22cm

The perimeter of the wire = Perimeter of the rectangle = 2 (l + b) = 2 x (40 cm + 22 cm) = 2 x 62 cm = 124 cm

Then, the perimeter of the square = 124 cm

The side length of the square = (124/4) cm = 31 cm

Area of the rectangle = l x b = 40 cm x 22 cm = 880 cm^2

Area of the square = s^2 = 31 cm x 31 cm = 961 cm^2

The shape which encloses more are = Area of square - area of reactangle

=>961cm^2-880cm^2
=>81cm^2

This proves that the square encloses more area than the rectangle.

By the way, whenever you have a certain perimeter length, the largest quadrilateral that has this perimeter is always a square....

Answer 2

Hey there !

Solution:

Length of wire = 40 cm

Breadth of wire = 22 cm

=> Perimeter of Wire = 2 ( l + b )

=> Perimeter = 2 ( 40 + 22 )

=> Perimeter = 2 ( 62 ) = 124 cm

Hence perimeter of wire in the shape of rectangle is 124 cm.

Now it is in the form of square. So let us assume the side to be 'a'.

=> Perimeter of Square = Perimeter of Rectangle

=> 4 a = 124

=> a = 124 / 4 = 31

Hence one side of a square is 31 cm.

Area of Rectangle = l × b

=> Area of Rectangle = 40 × 22 = 880 cm²

Area of Square = a²

=> Area of Square = 31 × 31 = 961 cm²

So the area covered by square is more than that of area covered by rectangle.

Area Difference = 961 - 880 = 81 cm²

Hope my answer helped !