If the area of a rectangular field is 2a2 +7ab -15b2
sq. m, find the dimensions of the field
in terms of the variables.
Answers
Answer:
Given−
\rm :\longmapsto\:Area_{(rectangle)} = {2a}^{2} + 7ab - {15b}^{2}:⟼Area
(rectangle)
=2a
2
+7ab−15b
2
\begin{gathered}\begin{gathered}\begin{gathered}\bf \: To \: Find - \begin{cases} &\sf{Breadth \:} \\ &\sf{Length } \end{cases}\end{gathered}\end{gathered}\end{gathered}
ToFind−{
Breadth
Length
\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}} \end{gathered}
FormulaUsed−
1. \: \: \: \boxed{ \sf \: Area_{(rectangle)} = Length \times Breadth}1.
Area
(rectangle)
=Length×Breadth
\large\underline{\sf{Solution-}}
Solution−
Given that
\rm :\longmapsto\:Area_{(rectangle)} = {2a}^{2} + 7ab - {15b}^{2}:⟼Area
(rectangle)
=2a
2
+7ab−15b
2
\rm :\longmapsto\:Area_{(rectangle)} = {2a}^{2} + 10ab - 3ab - {15b}^{2}:⟼Area
(rectangle)
=2a
2
+10ab−3ab−15b
2
\rm :\longmapsto\:Area_{(rectangle)} = {2a}(a + 5b) - 3b (a + {5b} ):⟼Area
(rectangle)
=2a(a+5b)−3b(a+5b)
\bf\implies \:Area_{(rectangle)} = (2a - 3b)(a + 5b)⟹Area
(rectangle)
=(2a−3b)(a+5b)
As we know,
\rm :\longmapsto\:Area_{(rectangle)} = Length \times Breadth:⟼Area
(rectangle)
=Length×Breadth
So, on comparing, we get
\begin{gathered}\begin{gathered}\begin{gathered}\bf \: Dimensions - \begin{cases} &\sf{Breadth = (2a - 3b) \:units} \\ &\sf{Length = (a + 5b) \: units} \end{cases}\end{gathered}\end{gathered}\end{gathered}
Dimensions−{
Breadth=(2a−3b)units
Length=(a+5b)units
Or
\begin{gathered}\begin{gathered}\begin{gathered}\bf \: Dimensions- \begin{cases} &\sf{Breadth = (a + 5b)\:units} \\ &\sf{Length = (2a - 3b) \: units} \end{cases}\end{gathered}\end{gathered}\end{gathered}
Dimensions−{
Breadth=(a+5b)units
Length=(2a−3b)units
Additional Information :-
1. \: \: \: \boxed{ \sf \: Perimeter_{(rectangle)} = 2(Length + Breadth)}1.
Perimeter
(rectangle)
=2(Length+Breadth)
2. \: \: \: \boxed{ \sf \: Area_{(circle)} = \pi \: {r}^{2} }2.
Area
(circle)
=πr
2
3. \: \: \: \boxed{ \sf \: Perimeter_{(circle)} = 2\pi \: r}3.
Perimeter
(circle)
=2πr
4. \: \: \: \boxed{ \sf \: Perimeter_{(square)} = 4 \times side}4.
Perimeter
(square)
=4×side
5. \: \: \: \boxed{ \sf \: Area_{(square)} = {(side)}^{2} }5.
Area
(square)
=(side)
2