Question

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Answer 1

• Understanding the concept :

❍ To find the area of square and the rectangle first we have to find the 4s of square. We can calculate the breadth of the rectangle using the formula of the perimeter of rectangle. At the last we will find that whose area is greater.

　　　　　　　　　　　 

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• S O L U T I O N :

　　　　　　　　　　　 

Side of the square = 10 cm

Length of the wire = Perimeter of square

Perimeter of square = 4 × Side = 4 × 10 cm

　　　　　= 40 cm

Length of the rectangle, ℓ = 12 cm.

( Let b be the Breadth of rectangle )

Perimeter of the rectangle = Length of wire

　　　　　= 40 cm

Perimeter of the rectangle = 2 (ℓ + b)

Thus, 　　　　　40 = 2 (12 + b)

Therefore,　　　b = 20 – 12 = 8 cm

The Breadth of rectangle is 8 cm

　　　　　 

Area of the square = (side)²

　　　　　 = 10 cm × 10 cm = 100 cm²

　　　　　 　　　　　 

Area of the rectangle = ℓ × b

= 12 cm × 8 cm = 96 cm²

So, The square encloses more area even though its perimeter is the same of the rectangle.

Answer 2

Provided: Length of the square is 10cm & length of the rectangle is 12cm.

Need to find: Breadth of the rectangle & which encloses more area.

❍ Let b be the breadth of the rectangle.

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$\underline{\bigstar\;\frak{According\;to\;the\;Question:}}$

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• The wire which is in the shape of a square, is rebent into a rectangle.

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Therefore,

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$:\implies\sf{Perimeter\;_{(square)}=Perimeter\;_{(rectangle)}}\\\\\\\\:\implies\sf{4\times Side=2(Length+Breadth)}\\\\\\\\:\implies\sf{4\times 10=2(12+b)}\\\\\\\\:\implies\sf{40=2(12+b)}\\\\\\\\:\implies\sf{\cancel{\dfrac{40}{2}}=12+b}\\\\\\\\:\implies\sf{20=12+b}\\\\\\\\:\implies\sf{20-12=b}\\\\\\\\:\implies\underline{\boxed{\pink{\frak{b=8\;cm}}}}{\;\bigstar}$

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$\therefore\;{\underline{\sf{Hence,\;the\;breadth\;of\;the\;rectangle\;is\;{\textsf{\textbf{8\;cm}}}}.}}$

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Now, we've to calculate the area of the square and the rectangle to find which encloses more area.

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✭ Area of the square:

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$:\implies\sf{(Side)^2}\\\\\\:\implies\sf{(10)^2}\\\\\\:\implies\frak{100\;cm^2}$

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✭ Area of rectangle:

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$:\implies\sf{(Length\times Breath)}\\\\\\:\implies\sf{12\times 8}\\\\\\:\implies\frak{96\;cm^2}$

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$\therefore\;{\underline{\sf{Hence,\;the\;{\textsf{\textbf{square}}}\;encloses\;more\;area.}}}$⠀⠀⠀