Question

(22) A wire is in the shape of side 12 cm, if the wire is rebent into a rectangle of length 14 cm, find its breadth. Which encloses more area, the square the rectangle?​

Answer 1

Answer :

• Area of square is greater by 4cm²

Given :

• A wire is in the shape of side is 12cm
• The wire is rebent into a rectangle of length 14cm

To find :

• Breadth
• which figure encloses more area

Solution :

Given that ,

• Side of a square = 12cm

As we know that,

• Perimeter of square = 4 × s

》4 × 12 = 48cm

Hence, Perimeter of square is 48cm

Now ,

• Perimeter of square = Perimeter of rectangle

As we know that,

• Perimeter of rectangle = 2(length + breadth)

》2(l + b) = 48

》2(14 + b) = 48

》(14 + b) = 48/2

》14 + b = 24

》b = 24 - 14

》b = 10cm

Hence , Breadth of rectangle is 10cm

Now we have to find the which figure encloses more area :

• Area of square = (side)² = 12 × 12 = 144cm²
• Area of rectangle = length × breadth = 14 × 10 = 140cm²

》144 - 140 = 4

Hence , Area of square is greater by 4cm²

Answer 2

Answer:

GivEn:

Side of Square = 10 cm

Length of Rectangle = 12 cm

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To find:

Which encloses more area, the square or the rectangle?

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Solution:

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A wire of Square shape is rebent into Rectanglular shape.

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We have,

Side of Square = 10 cm

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We know that,

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$\star\;{\boxed{\sf{\purple{Perimeter_{\;(square)} = 4 \times side}}}}$

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$:\implies\sf Perimeter_{\;(square)} = 4 \times 10$

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$:\implies\bf Perimeter_{\;(square)} = 40\;cm$

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Given that,

Length of Rectangle = 12 cm

Let breadth of Rectangle be b.

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Perimeter of square = Perimeter of Rectangle = 40 cm

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We know that,

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$\star\;{\boxed{\sf{\purple{Perimeter_{\;(Rectangle)} = 2(l + b)}}}}$

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$:\implies\sf Perimeter_{\;(Rectangle)} = 2(12 + b)$

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$:\implies\sf 40 = 2(12 + b)$

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$:\implies\sf \cancel{ \dfrac{40}{2}} = 12 + b$

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$:\implies\sf 20 = 12 + b$

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$:\implies\sf b = 20 - 12$

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$:\implies\bf b = 8\;cm$

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We know that,

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$\star\;{\boxed{\sf{\purple{Area_{\;(square)} = side \times side}}}}$

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☯ Therefore, Area enclosed by the square,

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$:\implies\sf Area_{\;(square)} = 10 \times 10$

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$:\implies\bf \pink{Area_{\;(square)} = 100\;cm^2}\;\bigstar$

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We know that,

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$\star\;{\boxed{\sf{\purple{Area_{\;(Rectangle)} = l \times b}}}}$

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☯ Therefore, Area enclosed by the square,

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$:\implies\sf Area_{\;(Rectangle)} = 12 \times 8$

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$:\implies\bf \pink{Area_{\;(Rectangle)} = 96\;cm^2}\;\bigstar$

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☯ Thus, The Difference in their area,

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$:\implies\sf Area_{\;(square)} - Area_{\;(Rectangle)}$

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$:\implies\sf 100 - 96$

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$:\implies\bf \blue{4\;cm^2}$

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$\therefore$ Hence, The square encloses more area by 4 cm².