Question

Wire is in the shape of a square of side 10 cm. If the wire is bent again into a rectangle of length 12 cm, find its breadth which encloses more are the square or the rectangle​

Answer 1

Answer:

Step-by-step explanation:

So,

Perimeter of the wire bent in square = 4* side

=4*10=40

Perimeter of square = perimeter of rectangle

Let be:

Breadth of the rectangular wire = x

Perimeter = 40

Perimeter = 2*( l+ b)

40. = 2*(12 + b)

40. = 24 + 2b

40 - 24. =. 2b

16. = 2b

16/2. = 8

Answer 2

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

A wire is in the shape of a square of side 10 cm. If the wire is rebent into a rectangle of length 12 cm, find its breadth. Which encloses more area, the square or the rectangle and how much.

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

$\sf Given\begin{cases} &\sf{Side\;of\;the\; square, \; a=\be{10\:cm}}\\&\sf{Length\;of\;the\;rectangle,\;l =\bf{12\;cm}}\end{cases}$

✏ 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 = 10 cm

⇒Perimeter = 4a

= 4 × 10

= 40 cm

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

• Length of the rectangle,l = 12 cm
• Breadth of the rectangle,b = ?

⇒Perimeter = 2 × (l + b)

= 2 × (12 + b)

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

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

⇒40 = 2 × (12 + b)

⇒(12 + b) = $\frac{40}{2}$

⇒b = 29 - 12

⇒b = 8 cm

$\pink{\underline\bold{ ∴ \:Breadth\:of\:the\:rectangle=\:8\: cm}}$

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

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

Area = a × a

= 10 × 10

= 100 cm2

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

Area = l × b

= 96 cm2

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

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