Question

The length of a rectangle is 12 m more than twice the width. The area of the rectangle is 320 spuare meter. Write an equation that can be used to find the lenght and width of the rectangle. Also find the dimension of the rectangle​

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \orange{Given :-}$

The length of a rectangle is 12 m more than twice the width.

The area of the rectangle is 320 m².

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \orange{To \: Find :-}$

Write an equation that can be used to find the length and width of the rectangle.

Formula Used :-

$\rightarrow$ Area Of Rectangle Formula :

$\footnotesize\mapsto \sf\boxed{\bold{\blue{Area_{(Rectangle)} =\: Length \times Breadth}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \bf \orange{Solution :-}$

Let,

$\mapsto \sf Width =\: a\: m$

$\mapsto \sf Length =\: (2a + 12)\: m$

According to the question by using the formula we get,

$\implies \sf 320 =\: (2a + 12) \times a$

$\implies \sf 320 =\: 2a^2 + 12a$

$\implies \sf 2a^2 + 12a - 320 =\: 0\: \: \bigg\lgroup \small\sf\bold{\pink{Required\: Equation}}\bigg\rgroup$

$\implies \sf 2a^2 + (32 - 20)a - 320 =\: 0$

$\implies \sf 2a^2 + 32a - 20a - 320 =\: 0\: \: \bigg\lgroup \small\sf\bold{\pink{By\: splitting\: middle\: term}}\bigg\rgroup$

$\implies \sf 2a(a + 16) - 20(a + 16) =\: 0$

$\implies \sf (a + 16)(2a - 20) =\: 0$

$\implies \bf a + 16 =\: 0$

$\implies \sf\bold{\purple{a =\: - 16}}\: \: \bigg\lgroup \small\sf\bold{\pink{Dimensions\: can't\: be\: negetive\: (- ve)}}\bigg\rgroup$

Either,

$\implies \sf 2a - 20 =\: 0$

$\implies \sf 2a =\: 20$

$\implies \sf a =\: \dfrac{\cancel{20}}{\cancel{2}}$

$\implies \sf\bold{\purple{a =\: 10}}$

Hence, the value of a = 10.

Hence, the required length and width are :

✩ Length Of Rectangle :

$\longrightarrow \sf Length_{(Rectangle)} =\: (2a + 12)\: m$

$\longrightarrow \sf Length_{(Rectangle)} =\: \{2(10) + 12\}\: m$

$\longrightarrow \sf Length_{(Rectangle)} =\: (20 + 12)\: m$

$\longrightarrow \sf\bold{\pink{Length_{(Rectangle)} =\: 32\: m}}$

✩ Width Of Rectangle :

$\longrightarrow \sf Width_{(Rectangle)} =\: x\: m$

$\longrightarrow \sf\bold{\pink{Width_{(Rectangle)} =\: 10\: m}}$

${\small{\bold{\underline{\therefore\: The\: length\: and\: width\: of\: rectangle\: are\: 32\: m\: and\: 10\: m\: respectively\: .}}}}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

Given : The length of a rectangle is 12m more than twice the width & the area of the rectangle is 320m².

To Find : Find the length and breadth of the rectangle ?

_________________________

Solution : Let the breadth be x and length be 2x.

$~$

$\underline{\frak{As ~we ~know ~that~:}}$

• $\boxed{\sf\pink{Area_{(Rectangle)}~=~ Length~×~ Breadth}}$

$~$

$\pmb{\sf{\underline{According~ to ~the ~Given~ Question~:}}}$

$~$

$\qquad{\sf:\implies{x~×~(2x~+~12)~=~320}}$

$\qquad{\sf:\implies{2x^2~+~12x~=~320}}$

$\qquad{\sf:\implies{2x^2~+~12x~-~320~=~0}}$

• Taking Common 2

$\qquad{\sf:\implies{x^2~+~6x~-~160~=~0}}$

$\qquad{\sf:\implies{x^2~+~16x~-~10x~-~160~=~0}}$

$\qquad{\sf:\implies{x(x~+~16)~- 10(x~+~6)~=~0}}$

$\qquad{\sf:\implies{(x~+~16)(x~-~6)~=~0}}$

$\qquad{\sf:\implies{x~+~16~=~0~~Or~~x~-~10~=~0}}$

$\qquad{\sf:\implies{\boxed{\frak{\purple{\pmb{x~=~- 16}}}}~~Or~~\boxed{\frak{\pink{\pmb{x~=~10}}}}}}$

$~$

Therefore,

• Since, dimension can't negative, So the value of x is 10m
• Length of the rectangle = 2 × 6 + 12 => 12 + 12 = 24m.

$~$

Hence,

$\therefore\underline{\sf{The ~length ~and ~breadth ~of~ the~ rectangle ~is~\bf{\underline{\pmb{32m}}}~\sf{\&}~\bf{\underline{\pmb{10m}}}}}$