Question

A wire is in the shape of a square of side 8 cm. If the wire is rebent into a rectangle of length 10 cm find its breadth. Which of the two shapes encloses larger area?

Answer 1

Hey Buddy

Here's The Answer

=> Firstly wire was in the shape of square of side 8 cm, so the perimeter of square will be the total length of wire

=> Length of wire = perimeter of square

=> Total length = 4 × side

=> Total length = 4 × 8

=> Total length = 32 cm

# Not the wire is rebent in the shape of rectangle of length ( l = 10 cm ) , we have to find it's breadth ( b )

=> total length = perimeter of rectangle

=> total length = 2 × ( l + b )

=> 32 = 2 × ( 10 + b )

=> 10 + b = 32 ÷ 2

=> 10 + b = 16

=> b = 16 - 10

=> breadth = 6 cm

# So the breadth of rectangle is 6 cm

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# Now we have to find which has larger area ,

=> Area of square = side × side

=> Area of square = 8 × 8

=> Area of square = 64 cm^2


# Now Area of rectangle = length × breadth

=> Area of rectangle = l × b

=> Area of rectangle = 10 × 6

=> Area of rectangle = 60 cm^2



# Hence Area of square ( 64 cm^2 ) is greater than area of rectangle ( 60 cm^2 )


HOPE HELPED.

JAI HIND

PEACE

:)

Answer 2

________Heyy Buddy ❤_______

______Here's your Answer ________

Given :-

Side of Square = 8cm

=> Perimeter of Square = 4 × 8

=> Perimeter = 32 cm.

Now,

Length of Rectangle = 10 cm.

Since, The square wire is bent in the form of Rectangle.

Therefore, The Perimeter of Square wil b eequal to the Perimeter of Rectangle.

=> Perimeter of Square = Perimeter of Rectangle

=> 32 = 2 ( l + b)

=> 10 + b = 16

=> b = 6 cm.

Now,


Area of Square = side × side

=> 8 × 8

=> 64 cm^2.

And Area of Rectangle = Length × Breadth

=> 10 × 6

=> 60 cm^2.

Since, Area of Square is greater than area of Rectangle.

Therefore,

Area of square encloses the larger area.