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
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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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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