Question

The area of rectangle is 132cm . They have given the breadth 6 cm . What is the length?

Answer 1

Given :

• Area of rectangle = 132 cm²
• Breadth of rectangle = 6 cm

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To find :

• Length of Rectangle?

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Solution :

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☯ Let length of Rectangle be x cm.

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$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large x cm}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large 6 cm}\put(-0.5,-0.4){\bf A}\put(-0.5,3.2){\bf D}\put(5.3,-0.4){\bf B}\put(5.3,3.2){\bf C}\end{picture}$

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We know that,

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$:\implies{\boxed{\sf{\purple{Area_{\;(Rectangle)} = length \times breadth}}}}$

Now, Putting values,

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$:\implies\sf x \times 6 = 132$

$:\implies\sf x = \cancel{ \dfrac{132}{6}}$

$:\implies{\boxed{\sf{\pink{x = 22\;cm}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\; Length\;of\;rectangle\;is\; \bf{22\;cm}.}}}$

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$\qquad\qquad\boxed{\underline{\underline{\bigstar \: \bf\:More\:to\:know\:\bigstar}}}$

• Perimeter of Rectangle = 2(l + b)

• Diagonal of Rectangle = $\sf \sqrt{l^2 + b^2}$

• Area of square = side × side

• Perimeter of square = 4 × side

• Diagonal of square = $\sf a \sqrt{2}$

Answer 2

Question :

To find length of Rectangle

Solution :

Given ::

• Area of rectangle = 132 cm²
• Breadth = 6 cm

${\underline{\boxed{\bf{area \: of \: Rectangle \: = l \times b \: \: }}}}$

• l = length
• b = breadth

Let's do ::

let, Length be x ,

$\sf \longrightarrow \: x \: \times 6 = 132 \\ \\\sf\longrightarrow \: x = \frac{132}{6} \\ \\ \bf\longrightarrow \: x = 22$

So, Length is 22 cm.

let's verify ::

$= > \sf \: l \: \times b \: = 132$

$= > \sf \: 22 \times 6 = 132$

$= > \sf \: 132 = 132$

$\bf \: LHS \: = \: RHS \:$

Thus Solved !!