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

Answer :-

⑅ The breadth of the rectangle = 10m

Given :-

✧ The area of the rectangle (l) = 320m²

To find :-

◍ The breadth of the rectangle

Solution :-

Let the breadth be x.

Then, length = 12m + 2x

In the question, the length and area of rectangle is given and we have to calculate the dimensions of rectangle.

Area of rectangle is the product of length and base.

$\mapsto \: \tt{Area \: \: of \: \: rectangle= Base \times Length}$

Substituting the values we have,

$\mapsto \tt{320 \:m^2= (12+2x) \times x}\\\\\mapsto \tt{320 \:m^2= 2(6+x) \times x}\\\\\mapsto \tt{ \dfrac{320 \:m^2}{2} = (6+x) \times x}\\\\\mapsto \tt{160 \:m^2= 6x+ {x}^{2}}\\\\\mapsto \tt{0= 6x+ {x}^{2} - 160 \:{m}^{2} }\\\\\mapsto \tt{0={x}^{2} + 6x- 160}\\\\\mapsto \tt{0={x}^{2} - 10x+ 16x- 160}\\\\\mapsto \tt{0=x(x - 10)+ 16(x- 10)}\\\\\mapsto \tt{0=(x - 10)(x + 16)}\\\\\mapsto \tt{x = 10 \: \: or \: \: - 16}$

As, breadth cannot be negative. Therefore, x = 10m.

Length of the rectangle

= 12m + 2 × x

= 12m + 2 × 10m

= 12m + 20m

= 32m

More Information :-

• Area of square = (Side)²

• Area of rectangle = Length × Breadth

• Perimeter of square = 4 × side

• Perimeter of rectangle = 2(Length + Breadth)

• Area of rhombus = 1/2 × First Diagonal × Second Diagonal

• Perimeter of rhombus = 4 × Side

• Area of parallelogram = Base × Height

Answer 2

Given:—

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

• The area of the rectangle is 320 m².

To Find:—

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

Formula Used:—

$\clubsuit♣ \: Area \: Of \: Rectangle \: Formula :$

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

Solution:—

Let,

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

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

According to the question by using the formula we get,

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

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

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

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

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

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

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

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

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

Either,

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

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

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

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

Hence,

→ The value of a = 10.

Hence,

→ The required length and width are :

❒ Length Of Rectangle:—

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

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

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

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

❒ Width Of Rectangle:—

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

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

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