Question

A wire is in the shape of a square of side 12 cm. It is again bent into a rectangle of length 15 cm. Find it breadth which encloses more area and how much.​

Answer 1

Answer:

SQUARE

Given side of square, s = 12cm

Perimeter = 4s = 4 x 12 = 48 cm

Area = s² = 12² = 144 cm²

RECTANGLE

Given length, l = 15cm

breadth, b = ?

As the same piece of wire is used to create the rectangle, perimeter of both, the square and the rectangle will remain the same.

∴ 48 = 2 (l + b)

48/2 = 15 + b

24 = 15 + b

24 - 15 = b

b = 9 cm

Area = l x b

= 15 x 9

= 135 cm²

Area of square = 144cm²

Area of rectangle = 135cm²

∴ The square occupies more area by 9cm²

Hope it helps you :)

Answer 2

$\huge\rm\underline\purple{Question :-}$

A wire is in the shape of a square of side 12 cm. It is again bent into a rectangle of length 15 cm. Find it breadth which encloses more area and how much.

$\huge\rm\underline\purple{Answer :-}$

$\sf Given\begin{cases} &\sf{Side\;of\;the\; square, \; a=\be{12\:cm}}\\&\sf{Length\;of\;the\;rectangle,\;l =\bf{15\;cm}}\end{cases}$

✏ Here, the square is bent into a rectangle. So, the length of the wire will be same.

$\red{\underline\bold{To\: find,}}$

• Breadth of the rectangle = ❓

$\green{\underline\bold{For\: the\:square,}}$

• Side of the square = a = 12 cm

⇒Perimeter = 4a

= 4 × 12

= 48 cm

$\green{\underline\bold{For\: the\:rectangle,}}$

• Length of the rectangle,l = 15 cm
• Breadth of the rectangle,b = ?

⇒Perimeter = 2 × (l + b)

= 2 × (15 + b)

$\blue{\underline\bold{Here,}}$

$\orange{\underline\bold{Perimeter\: of\:the\:square\: = \:Perimeter\: of\: the\: rectangle}}$

⇒48 = 2 × (15 + b)

⇒(15 + b) = $\frac{48}{2}$

⇒b = 24 - 15

⇒b = 9 cm

$\pink{\underline\bold{ ∴ \:Breadth\:of\:the\:rectangle=\:9\: cm}}$

$\green{\underline\bold{For\:finding\:the\:area,}}$

$\blue{\underline\bold{For\: the\:square,}}$

Area = a × a

= 12 × 12

= 144 cm2

$\blue{\underline\bold{For\: the\:rectangle,}}$

Area = l × b

= 135 cm2

$\purple{\underline\bold{So,\: area\:of\: the\: square \:is \:greater\: than \:area \:of \:a\: rectangle.}}$

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