Question

If the area of a rectangular field is 2a^2 +7ab -15b^2 sq. m, find the dimensions of the field in terms of the variables.

Answer 1

$\large\underline{\sf{Given- }}$

$\rm :\longmapsto\:Area_{(rectangle)} = {2a}^{2} + 7ab - {15b}^{2}$

$\begin{gathered}\begin{gathered}\bf \: To \: Find - \begin{cases} &\sf{Breadth \:} \\ &\sf{Length } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$1. \: \: \: \boxed{ \sf \: Area_{(rectangle)} = Length \times Breadth}$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:Area_{(rectangle)} = {2a}^{2} + 7ab - {15b}^{2}$

$\rm :\longmapsto\:Area_{(rectangle)} = {2a}^{2} + 10ab - 3ab - {15b}^{2}$

$\rm :\longmapsto\:Area_{(rectangle)} = {2a}(a + 5b) - 3b (a + {5b} )$

$\bf\implies \:Area_{(rectangle)} = (2a - 3b)(a + 5b)$

As we know,

$\rm :\longmapsto\:Area_{(rectangle)} = Length \times Breadth$

So, on comparing, we get

$\begin{gathered}\begin{gathered}\bf \: Dimensions - \begin{cases} &\sf{Breadth = (2a - 3b) \:units} \\ &\sf{Length = (a + 5b) \: units} \end{cases}\end{gathered}\end{gathered}$

Or

$\begin{gathered}\begin{gathered}\bf \: Dimensions- \begin{cases} &\sf{Breadth = (a + 5b)\:units} \\ &\sf{Length = (2a - 3b) \: units} \end{cases}\end{gathered}\end{gathered}$

Additional Information :-

$1. \: \: \: \boxed{ \sf \: Perimeter_{(rectangle)} = 2(Length + Breadth)}$

$2. \: \: \: \boxed{ \sf \: Area_{(circle)} = \pi \: {r}^{2} }$

$3. \: \: \: \boxed{ \sf \: Perimeter_{(circle)} = 2\pi \: r}$

$4. \: \: \: \boxed{ \sf \: Perimeter_{(square)} = 4 \times side}$

$5. \: \: \: \boxed{ \sf \: Area_{(square)} = {(side)}^{2} }$