Question

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

Answer 1

side of square= 12 cm

perimeter= l of wire

p= 4*side

=4*12

l of wire =48 cm

perimeter of rectangle= l of wire

2(l+b)= 48

15 +b= 48/2

b=24-15

breadth= 9 cm-----------------1

area of rectangle=

l*b

=135 cm square----------------2

area of square=

side*side

=12*12

=144 cm square-----------------3

hence the square has more area------------------4

Done!!

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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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