Question

A wire is in the shape of a rectangle. Its length is 36cm and
breadth is 32cm. If the same wire is bent in the shape of a
square, what will be the measure of each side? Also find
which shape encloses more area?

Answer 1

Step-by-step explanation:

area of rectangle = length and width

= 36*32

=1152 cm2

Area of square = side *side

let side of square be x

x*x =1152

x2 = 1152 cm2

Answer 2

$\tt \pink{Given}\begin{cases}&\sf{Length\:of\:the\:rectangle=\bf{36\:cm.}} \\ \\ &\sf{Breadth\:of\:the\:rectangle=\bf{32\:cm.}}\end{cases}$

To FinD:-

• Measure of each side of square.
• Shape that encloses more area.

SolutioN:-

We know that,

$\normalsize\quad{\purple{\underline{\boxed{\sf{Perimeter_{(rectangle)}=2(Length+Breadth)}}}}}$

$\tt {Here}\begin{cases}&\sf{Length=\bf{36\:cm}} \\ \\ &\sf{Breadth=\bf{32\:cm}}\end{cases}$

• Putting the values,

$\\ :\normalsize\implies{\sf{Perimeter_{(Rectangle)}=2(36+32)}}$

$\\ \qquad:\normalsize\implies{\sf{Perimeter_{(Rectangle)}=2\times68}}$

$\\ \qquad\quad\quad\normalsize\therefore\boxed{\mathfrak{\pink{Perimeter_{Rectangle)}=136\:cm.}}}$

Now,

• It is said that the wire is bent in the shape of a square.

$\\ \ \ \ \ \ \normalsize{\sf{\pink{\underline{Perimeter_{(Rectangle)}=Perimeter_{(square)}}}}}$

We know that,

$\normalsize\qquad\quad{\purple{\underline{\boxed{\sf{Perimeter_{(square)}=4\times side}}}}}$

$\tt {Here}\begin{cases}&\sf{Perimeter_{(square)}=\bf{136\:cm}} \\ \\ &\sf{Side=\bf{?}}\end{cases}$

• Putting the values,

$\\ :\normalsize\implies{\sf{136=4\times side}}$

$\\ \qquad:\normalsize\implies{\sf{\dfrac{136}{4}=side}}$

$\\ \qquad\quad:\normalsize\implies{\sf{\cancel{\dfrac{136}{4}}=side}}$

$\\ \qquad\quad\quad\normalsize\therefore\boxed{\mathfrak{\pink{Side_{(square)}=34\:cm.}}}$

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Area of rectangle :

We know that,

$\normalsize\qquad\quad{\purple{\underline{\boxed{\sf{Area_{(rectangle)}=Length\times Breadth}}}}}$

$\tt {Here}\begin{cases}&\sf{Length=\bf{36\:cm}} \\ \\ &\sf{Breadth=\bf{32\:cm}}\end{cases}$

• Putting the values,

$\\ :\normalsize\implies{\sf{Area_{(Rectangle)}=36\times32}}$

$\\ \qquad\quad\quad\normalsize\therefore\boxed{\mathfrak{\pink{Area_{(Rectangle)}=1152\:cm^2.}}}$

Area of square :

We know that,

$\normalsize\qquad\quad{\purple{\underline{\boxed{\sf{Area_{(square)}=side\times side}}}}}$

$\tt {Here}\begin{cases}&\sf{Side=\bf{34\:cm}}\end{cases}$

• Putting the values,

$\\ :\normalsize\implies{\sf{Area_{(square)}=34\times34}}$

$\\ \qquad\quad\quad\normalsize\therefore\boxed{\mathfrak{\pink{Area_{(square)}=1156\:cm^2.}}}$

By comparing the area of both rectangle and square,

• Square encloses more area.

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• Measure of each side of square is 34 cm.
• Square encloses more area.