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In the centre of a rectangular lawn of dimension $50\ m \times 40\ m$, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be $1184\ m^2$. Find the length and breadth of the pond.
Given:
Dimensions of the rectangular lawn are $50\ m \times 40\ m$.
Area of the grass surrounding the pond$=1184\ m^2$.
To do:
We have to find the length and breadth of the pond.
Solution:
Area of the rectangular lawn$=50\ m \times 40\ m=2000\ m^2$.
Area of the pond$=$Area of the lawn$-$Area of the grass
Area of the pond$=(2000-1184)\ m^2=816\ m^2$.
Let the width of the grass surrounding the pond be $x\ m$.
This implies,
Length of the pond$=50-2x\ m$.
Breadth of the pond$=40-2x\ m$.
We know that,
Area of a rectangle of length $l$ and breadth $b$ is $lb$.
Therefore,
Area of the rectangular pond$=(50-2x)(40-2x)\ m^2$.
According to the question,
$(50-2x)(40-2x)=816$ (From equation 1)
$2000-100x-80x+4x^2=816$
$4x^2-180x+2000-816=0$
$4x^2-180x+1184=0$
$4(x^2-45x+296)=0$
$x^2-45x+296=0$
Solving for $x$ by factorization method, we get,
$x^2-37x-8x+296=0$
$x(x-37)-8(x-37)=0$
$(x-37)(x-8)=0$
$x-37=0$ or $x-8=0$
$x=37$ or $x=8$
If $x=37$, length of pond$=50-2(37)=50-74=-24$, which is not possible.
Therefore, the value of $x=8$.
$50-2x=50-2(8)=50-16=34\ m$
$40-2x=40-2(8)=40-16=24\ m$
The breadth of the pond is $24\ m$ and the length of the pond is $34\ m$.