Question

The ratio between length and breadth of a field is 10 : 6. The area of the field is 3840m². Find the difference between the length and width of the field.​

Answer 1

Given : The ratio between length and breadth of a field is

10 : 6. The area of the field is 3840m²

To Be Found: the difference between the length and width of the field.

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❒ Let the Length of the Rectangle be 10x and the breadth of the rectangle be 6x

${ \underline{ \cal{ \bold{ \bigstar \: According \: to \: the \: question : }}}}$

The length and the breadth are in the ratio 10 : 6 and the area of the field is 3840m² respectively!

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

$\: \: \: \: \: \: \: \: \: \: \: { \dag{ \bf{ \bigg(Area \: of \: a \: rectangle = l \times b \bigg)}}}$

★Where,

L stands for length

B stands for breadth

${ \bf{ \underline{ \circ \: subtitutuing \: the \: values : }}}$

${ : \implies} \sf \: area_{(rectangle)} = length \times breadth \\ \\ \\ { : \implies} \sf3840 {m}^{2} = 10x \times 6x \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \sf3840 {m}^{2} = 60 {x}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \sf \: {x}^{2} = \cancel\frac{3840}{60} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \sf \: {x}^{2} = 64\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \sf \: x = \sqrt{64} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ { : \implies} \sf \: { \purple{ \boxed{ \frak{x = 8m}} \star}}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Now,

let's find the length and breadth of the rectangle respectively

Here,

Length = 10x = 80m

Breath = 6x = 48m

$\: \: \: { \underline{ \rm{ \therefore \: the \: length \: and \: breadth \: of \: the \: field \: are \: 80m \: and \: 48m}}}$

Now,

Let's find the difference between the length and breadth of the rectangle!

$\dashrightarrow \sf \: difference = lenght - breadth \\ \\ \\ \dashrightarrow \sf \: difference = 80m - 48m\: \: \: \: \: \: \: \: \: \: \\ \\ \\ \dashrightarrow \sf \: difference = { \pink{ \boxed{ \frak{32m}}}}\bigstar \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

${ \boxed{ \boxed{ \rm{ \therefore \: the \: difference \: between \: the \: lenght \: and \: breadth = \red{32m}}}}}$

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$\: \: \: \: \: \: \: \: \: \: \: { \bf{ \overline{ \mid{ \underline{more \: to \: know \mid}}}}}$

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