Question

The lenght of a rectangle is 12 m more than twice the width .The area of the rectangle is 320 square m. 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

Given
length of a rectangle is 12m more than twice the width
Area of the rectangle is 320 m
let the width of a rectangle be = x
then the length of a rectangle = 2x + 12 (given)
we know that ,
area of a rectangle = l×b
x (2x +12) =320 (given)
2x ^2 + 12 x = 320
2x^2 + 12x - 320 =0 is the required equation

Answer 2

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

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 :

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

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

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