Appropriate Question :-

The perimeter of a rectangle is 32 cm. If its length is 2 cm more than its width, find its area.

[tex]\large\underline{\sf{Solution-}}[/tex]

Given that, length is 2 cm more than its width.

Let assume that

Width of a rectangle = x cm

Length of a rectangle = x + 2 cm

Further given that, Perimeter of a rectangle is 32 cm.

We know,

[tex]\boxed{ \sf{ \:Perimeter_{(Rectangle)} = 2(length \: + \: width )\: }} \\ \\ [/tex]

So, on substituting the values, we get

[tex]\sf \: 2(x + 2 + x) = 32 \\ \\ [/tex]

[tex]\sf \: 2(2x + 2) = 32 \\ \\ [/tex]

[tex]\sf \: 4x + 4 = 32 \\ \\ [/tex]

[tex]\sf \: 4x = 32 - 4\\ \\ [/tex]

[tex]\sf \: 4x = 28\\ \\ [/tex]

[tex]\bf\implies \:x = 7 \: cm \\ \\ [/tex]

So,

Width of a rectangle = x = 7 cm

Length of a rectangle = x + 2 = 7 + 2 = 9 cm

Now, We know

[tex]\boxed{ \sf{ \:Area_{(Rectangle)} = length \: \times \: width \: }} \\ \\ [/tex]

So, on substituting the values, we get

[tex]\sf \: Area_{(Rectangle)} = 9 \times 7 \\ \\ [/tex]

[tex]\bf\implies \:Area_{(Rectangle)} = 63 \: {cm}^{2} \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}[/tex]

Answer:

Correct Question :-

• The perimeter of a rectangle is 32 cm. If its length is 2 cm more than its width. Find its area.

Given :-

• The perimeter of a rectangle is 32 cm.
• It's length is 2 cm more than its width.

To Find :-

• What is the area of rectangle.

Solution :-

First, we have to find the length and width of a rectangle :

Let,

[tex]\mapsto \bf Width_{(Rectangle)} =\: x\: cm\\[/tex]

[tex]\mapsto \bf Length_{(Rectangle)} =\: (x + 2)\: cm\\[/tex]

Given :

• Perimeter of Rectangle = 32 cm

According to the question by using the formula we get,

[tex]\small \implies \sf\boxed{\bold{Perimeter_{(Rectangle)} =\: 2(Length + Width)}}\\[/tex]

[tex]\implies \sf 32 =\: 2\bigg\{x + (x + 2)\bigg\}\\[/tex]

[tex]\implies \sf 32 =\: 2(x + x + 2)[/tex]

[tex]\implies \sf 32 =\: 2(2x + 2)[/tex]

[tex]\implies \sf \dfrac{32}{2} =\: 2x + 2[/tex]

[tex]\implies \sf 16 =\: 2x + 2[/tex]

[tex]\implies \sf 16 - 2 =\: 2x[/tex]

[tex]\implies \sf 14 =\: 2x[/tex]

[tex]\implies \sf \dfrac{14}{2} =\: x[/tex]

[tex]\implies \sf 7 =\: x[/tex]

[tex]\implies \sf\bold{x =\: 7}[/tex]

Hence, the required length and width of a rectangle will be :

[tex]\dag[/tex] Length Of Rectangle :

[tex]\dashrightarrow \sf Length_{(Rectangle)} =\: (x + 2)\: cm\\[/tex]

[tex]\dashrightarrow \sf Length_{(Rectangle)} =\: (7 + 2)\: cm\\[/tex]

[tex]\dashrightarrow \sf\bold{Length_{(Rectangle)} =\: 9\: cm}\\[/tex]

[tex]\dag[/tex] Width Of Rectangle :

[tex]\dashrightarrow \sf Width_{(Rectangle)} =\: x\: cm\\[/tex]

[tex]\dashrightarrow \sf\bold{Width_{(Rectangle)} =\: 7\: cm}\\[/tex]

Now, we have to find the area of a rectangle :

Given :

• Length of Rectangle = 9 cm
• Width of Rectangle = 7 cm

According to the question by using the formula we get,

[tex]\small \longrightarrow \sf\boxed{\bold{Area_{(Rectangle)} =\: Length \times Width}}\\[/tex]

[tex]\longrightarrow \sf Area_{(Rectangle)} =\: 9\: cm \times 7\: cm\\[/tex]

[tex]\longrightarrow \sf\bold{\underline{Area_{(Rectangle)} =\: 63\: cm^2}}\\[/tex]

[tex]\therefore[/tex] The area of a rectangle is 63 cm² .