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

$\huge{\bf{\green{\mathfrak{\dag{\underline{\underline{Question:-}}}}}}}$

$\bullet{\longmapsto}$ 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.

$\huge {\bf{\orange{\mathfrak{\dag{\underline{\underline{Answer:-}}}}}}}$

$\bullet{\leadsto} \: \textsf{The ratio between length and breadth of a field is 10:6.}$

$\bullet{\leadsto} \: \sf Area \: of \: the \: field \: = \: 3840m^{2}.$

$\bullet{\leadsto} \: \textsf{Let the length and breadth of the field be x.}$

$\bullet{\leadsto} \: \pink{\tt Then, \: Length \: = \: 10x.}$

$\bullet{\leadsto} \: \pink{\tt Then, \: Breadth \: = \: 6x.}$

$\bullet{\leadsto} \: \pink{\tt Area \: = \: 3840m^{2}.}$

$\bullet{\leadsto} \: \therefore{\sf l \: * \: b \: = \: 3840m^{2}.}$

$\bullet{\leadsto} \: \textsf{Substituting the values,}$

$\bullet{\leadsto} \: \sf 10x \: * \: 6x \: = \: 3840$

$\bullet{\leadsto} \: \sf 60x^{2} \: = \: 3840$

$\bullet{\leadsto} \: \sf x^{2} \: = \: \dfrac{384\cancel{0}}{6\cancel{0}}$

$\bullet{\leadsto} \: \sf x^{2} \: = \: \dfrac{\cancel{384}}{\cancel{6}}$

$\bullet{\leadsto} \: \sf x^{2} \: = \: 64$

$\bullet{\leadsto} \: \sf x \: = \: \sqrt{64}$

$\bullet{\leadsto} \: \underline{\boxed{\purple{\tt x \: = \: 8.}}}$

$\bullet{\leadsto} \: \textsf{Now finding length and breadth,}$

$\bullet{\leadsto} \: \sf Length \: = \: 10x \: = \: 10 \: * \: 8 \: = \: 80m.$

$\bullet{\leadsto} \: \sf Breadth \: = \: 6x \: = \: 6 \: * \: 8 \: = \: 48m.$

$\bullet{\leadsto} \: \pink{\texttt{Difference between length and breadth =}}$

$\bullet{\leadsto} \: \sf l \: - \: b$

$\bullet{\leadsto} \: \sf 80 \: - \: 48$

$\bullet{\leadsto} \: \underline{\boxed{\purple{\tt Difference \: = \: 32 \: m.}}}$

$\huge{\bf{\red{\mathfrak{\dag{\underline{\underline{Conclusion:-}}}}}}}$

$\bullet{\longmapsto} \: \boxed{\therefore{\sf Difference \: between \: length \: and \: breadth \: of \: the \: field \: = \: 32 \: m.}}$

$\huge{\bf{\blue{\mathfrak{\dag{\underline{\underline{Formulas \: Used:-}}}}}}}$

$\bullet{\leadsto} \: \underline{\sf Area \: of \: Rectangle \: = \: l \: * \: b.}$

$\bullet{\leadsto} \: \sf where,$

$\bullet{\leadsto} \: \sf l \: is \: Length$

$\bullet{\leadsto} \: \sf b \: is \: Breadth$

Answer 2

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!

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

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

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

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

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