Question

A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is bent in the shape of a square, what will be the measure of each side. Also, find which side encloses more area?

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Answer 1

★Answer :-

• Square shape wire enclose more area .

★ Given :-

• A wire is in the Shape of rectangle.
• The Length of the rectangle is 40 cm .
• Breadth of the rectangle is 22 cm .

$\bigstar \underline {\rm \bf Solution }:-$

As we know , Same Length of wire will be there only . Whatever Shape we make it . So , same length of wire will be there if we make it square or rectangle .

That means :

$\sf {Perimeter \: of \: Square \: = \: Perimeter \: of \: rectangle} \: \star$

We know that :-

$\rm Perimeter \: of \: Square = \bf4 \times side \: \star$

$\rm \: Perimeter \: of \: rectangle = \bf \: 2(length + breadth) \: \star$

Let's solve :-

$\sf {Perimeter \: of \: Square \: = \: Perimeter \: of \: rectangle} \: \star$

$\sf \implies 4 \times side = 2(length + breadth)$

Only length and breadth is provided so , let's put the value

$\sf \implies 4 \times side = 2(40 + 22)$

$\sf \implies 4 \times side =2(62)$

$\sf \implies 4 \times side =2 \times 62$

$\sf \implies 4 \times side = 124$

Side remains this side but 4 will be divided

$\sf \implies side = \dfrac{124}{4}$

$\sf \implies side = 31 \: cm \: \star$

★Hence , side is 31 cm .

Now we need to find out which shape encloses more area ,

$\rm \: \bf \: Area \: of \: Square = side \times side\star$

$\rm \: \implies \: Area \: of \: Square = ({31 \: cm}) ^{2}$

$\: \rm \: \implies \: Area \: of \: Square =\underline{ \bf {961 \: cm^{2}} }$

Now , also Area of rectangle is to found out

$\rm \bf \: Area \: of \: rectangle = length \times breadth\star$

$\rm \: \implies \: Area \: of \: rectangle = 40 \: cm \times 22 \: cm$

$\rm \: \implies \: Area \: of \: rectangle = \underline{\bf880 \: cm^{2} }$

Therefore , Square shape encloses more area .

★Additional Information !!

$\bigstar\rm\underline{\bf Formulae}:-$

• $\rm Area\:of\:Circle=\pi r^2$
• $\rm Perimeter\:of\:Circle=2\pi r$
• $\rm Area\:of\: Triangle=\dfrac12\times base\times height$
• $\rm Perimeter\:of\: Triangle=Sum\:of\:three\: sides(a+b+c)$
• $\rm Area\:of\: Square =a^2$
• $\rm Perimeter\:of\: Square =4a$
• $\rm Area\:of\: Rectangle=Length\times Breadth$
• $\rm Perimeter\:of\: Rectangle=2(Length+Breadth)$
• $\rm Area\:of\: Parallelogram = base\times height$
• $\rm Perimeter\:of\: Parallelogram =2(Side +Base)$

______________________________

Answer 2

★Answer:-

•Square shape will enclose more

area.

Given:-

• Length and breadth of a

rectangle.

Solution:-

As we know that the length of the wire will remain same either it will be a rectangle or a square.So, the perimeter of rectangle will be the same as square.

So, We will first find the perimeter of rectangle and will put the result in

formula of perimeter of square .

$\large \blue{\underline{ \green{ \boxed{\bf{ \red{Perimeter _{rectangle} = 2(l + b)}}}}}}$

Where,

• l = length

•b= breadth

$\large \blue{\underline{ \green{ \boxed{\bf{ \red{Perimeter_{square} = 4× side}}}}}}$

Where,

• s = side

Now we will find the perimeter of rectangle by substituting values we get:-

Perimeter of rectangle = 2(l+b)

=2(40cm+22cm)

= 2×62cm

= 124cm

Perimeter of square= 4× side

124cm= 4× side

124cm÷4= side

31cm=side

Hence the side of the square is 31cm.

Now we will find the area of both the figure to obtain the area enclosed by both figures.

$\large \blue{\underline{ \green{ \boxed{\bf{ \red{Area _{reactangle}= l×b}}}}}}$

Where,

•l = length

•b = breadth

Area of rectangle = l×b

=40cm×22cm

= 880cm²

$\large \blue{\underline{ \green{ \boxed{\bf{ \red{Area_{square}= {s}^{2} }}}}}}$

Where,

•s= side of square

Area of square= s²

= 31²

=961cm²

On comparing the are of rectangle and square we get to know that.

•880cm²<961cm²

As the area of square is greater. Henceforth,The area enclosed by square is greater than the area enclosed by square.