Question
In the centre of a rectangular lawn of dimensions $50\ m \times40\ m$, 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.
✓
Answer
Given that a rectangular pond has to be constructed in the centre of a rectangular lawn of dimensions $50\ m \times 40\ m.$
So, the distance between pond and lawn would be same around the pond. Say $x$ m.
Now, length of rectangular lawn $\left(l_1\right)=50 m$
and breadth of rectangular lawn $\left(b_1\right)=40 m$
Length of rectangular pond $(12)=50-(x+x)=50-2 x$
Also, area of the grass surrounding the pond $=1184 m^2$
Area of rectangular lawn - Area of rectangular pond = Area of grass surrounding the pond
$I_1 \times b_1-I_2 \times b_2=1184[\because \text { area of rectangle }=\text { length } x \text { breadth }]$
$\Rightarrow 50 \times 40-(50-2 x)(40-2 x)=1184$
$\Rightarrow 2000-\left(2000-80 x-100 x+4 x^2\right)=1184$
$\Rightarrow 80 x+100 x-4 x^2=1184$
$\Rightarrow 4 x^2-180 x+1184=0$
$\Rightarrow x^2-45 x+296=0$
$\Rightarrow x^2-37 x-8 x+296=0[\text { by splitting the middle term }]$
$\Rightarrow x(x-37)-8(x-37)=0$
$\Rightarrow(x-37)(x-8)=0$
$\Rightarrow x=8$
[At $x=37$, length and breadth, of pond are $-24$ and $-34$ , respectively but length and breadth cannot be negative. So, $x=37$ cannot be possible]
Length of pond $=50-2 x=50-2(8)=50-16=34 m$
and breadth of pond $=40-2 x=40-2(8)=50-16=34 m$
Hence, required length and .breadth of pond are $34\ m$ and $24\ m$ , respectively.
So, the distance between pond and lawn would be same around the pond. Say $x$ m.
Now, length of rectangular lawn $\left(l_1\right)=50 m$
and breadth of rectangular lawn $\left(b_1\right)=40 m$
Length of rectangular pond $(12)=50-(x+x)=50-2 x$
Also, area of the grass surrounding the pond $=1184 m^2$
Area of rectangular lawn - Area of rectangular pond = Area of grass surrounding the pond
$I_1 \times b_1-I_2 \times b_2=1184[\because \text { area of rectangle }=\text { length } x \text { breadth }]$
$\Rightarrow 50 \times 40-(50-2 x)(40-2 x)=1184$
$\Rightarrow 2000-\left(2000-80 x-100 x+4 x^2\right)=1184$
$\Rightarrow 80 x+100 x-4 x^2=1184$
$\Rightarrow 4 x^2-180 x+1184=0$
$\Rightarrow x^2-45 x+296=0$
$\Rightarrow x^2-37 x-8 x+296=0[\text { by splitting the middle term }]$
$\Rightarrow x(x-37)-8(x-37)=0$
$\Rightarrow(x-37)(x-8)=0$
$\Rightarrow x=8$
[At $x=37$, length and breadth, of pond are $-24$ and $-34$ , respectively but length and breadth cannot be negative. So, $x=37$ cannot be possible]
Length of pond $=50-2 x=50-2(8)=50-16=34 m$
and breadth of pond $=40-2 x=40-2(8)=50-16=34 m$
Hence, required length and .breadth of pond are $34\ m$ and $24\ m$ , respectively.
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