Question

a wire is in the shape of a square of perimeter 40 cm It is bent in the form of a rectangle of length 8 cm find the area of rectangle compare whose area is more the square or the rectangle​

Answer 1

Answer:

area of square is greater than area of rectangle for the solution refer the solution in the picture

Answer 2

$\huge {\underline {\sf \star \green{Question -}}}$

A wire is in the shape of a square of perimeter 40 cm. It is bent in the form of a rectangle of length 8 cm find the area of rectangle compare whose area is more the square or the rectangle.

$\huge{ \underline{ \sf \star { \green{Answer -}}}}$

$\underline{ \sf \red{Given -}}$

• Perimeter of square = 40 cm
• Length of rectangle = 8 cm

$\underline{ \sf \red{To \: find -}}$

• Whose area is more

$\underline { \sf \red{Solution -}}$

$\sf{Perimeter \: of \: square = Perimeter \: of \: rectangle}$

$\boxed{ \sf \purple{Perimeter \: of \: rectangle = 2(l+b)}} \\ \\ \sf \implies{40cm = 2(l+b)} \\ \sf \implies{40cm = 2(8 + b)} \\ \sf \implies{40cm = 16 + 2b} \\ \sf \implies{ 40 - 16 = 2b} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \implies{ 24 = 2b}$

$\sf \implies{ \frac{24}{12} = b} \\ \\ \sf \implies{2 = b} \\ \sf \implies{breadth = 2cm} \\ \\ \boxed{ \sf {\purple{ \implies{Area \: of \: rectangle = l × b}}}} \\ \\ \sf \implies{8 \times 2} \\ \sf \implies16 {cm}^{2}$

$\boxed{\sf {\implies { \purple{Perimeter \: of \: square = 4 × side}}}}$

$\sf \implies{40 = 4 \times s} \\ \sf \implies{ \frac{40}{4} = s } \\ \sf \implies{\frac{ \cancel{40} \: \: \: ¹⁰}{ \cancel{4}} = s} \\ \sf \implies{s = 10cm} \\ \\ \\ \boxed{ \sf{ \implies{ \purple{Area \: of \: square = s \times s}}}} \\ \\ \sf \implies{10 \times 10} \\ \sf \implies{20 \: {cm}^{2} }$

Therefore,

Area of square > Area of rectangle.