Question

A rectangle whose area is 96cm^2 is such that its length is 4metres longer than its width find. [a] it dimensions [b] its perimeter

Answer 1

Given

Area of the rectangle = 96 cm²

Length is 4 m longer than its width.

Let the width be x.

∴ Length = 4+x

$=> (4+x)(x)=96\\=> x^{2} +4x-96=0\\=>x^{2} +12x-8x-96=0\\=>x(x+12)-8(x+12)=0\\=> (x-8)(x+12)=0$

Now,

$x+12=0\\=> x = -12$

-ve value is not taken into account

∴ x = 8

=> Length = 4 + 8 = 12 m

    Breadth = 8 m

Perimeter

$2(length+breadth)\\=2(12+8)\\=2(20)$

= 40 m

                                                         

Answer 2

$\Large{\underline{\underline{\bf{Solution :}}}}$

We know the formula to find the area of rectangle

$\Large{\implies{\boxed{\boxed{\sf{Area = Length \times Breadth}}}}}$

Where,

• Breadth = a
• Length = 4a

__________________[Put Values]

$\sf{→x(x + 4) = 96} \\ \\ \sf{→x^2 + 4x = 96} \\ \\ \sf{x^2 + 4x - 96}$

$\sf{Factorise \: the \: Quadratic \: equation}$

$\begin{array}{crclc}&&-96x^2&&\\&&\swarrow\searrow&&\\&-8t&&12t&\end{array}\quad\boxed{\begin{minipage}{6cm}\item\bigstar \quad \; \, x^2+4x-96=0 \\\begin{itemize} \item x^2\times(-96)\;\,=-96x^2 \item -96x^2\qquad\;\, = 8x\times12x \item -8x+(12x)\, =4x \end{itemize}\end{minipage}}$

$\begin{aligned}\implies \sf{x^2+12x-8x-96}&=0\\\implies \sf{x(x+12)-8(x+12)}&=0\\\implies \mathsf{(x+12)(x-8)}&=0\\\implies \sf{x-8}&=0\\\implies \sf{t}&=8\end{aligned}$

$\sf\textsf{While equating the term (x + 12) to 0, }\\\sf\textsf{a negative value is obtained.}$

Length = x + 4 = 12 cm

Breadth = 8 cm

$\rule{150}{2}$

$\Large{\star{\boxed{\sf{Perimeter = 2(L + B)}}}}$

$\sf{→Perimeter = 2(12 + 8)} \\ \\ \sf{→Perimeter = 2(20)} \\ \\ \sf{→Perimeter = 40 \: cm}$