Question
A rectangular field is $20 m$ long and $14 m$ wide. There is a path of equal width all around it, having an area of $111\ sq\ m.$ Find the width of the path.
✓
Answer
Let the width of the path be $x m$
Length of the field including the path $= (20 + 2x) m$
Breadth of the field including the path $= (14 + 2x) m.$
Area of rectangle $= L B$
Area of the field including the path $= (20+2 x)(14+2 x) m ^2$.
Area of the field excluding the path $= (20 \times 14) m ^2=280 m^2$
$\therefore$ Area of the path $=(20+2 x)(14+2 x)-280$
$(20 + 2x) (14 + 2x) - 280 = 111$
$\Rightarrow 4 x^2+68 x-111=0$
Factorise the equation,
$\Rightarrow 4 x^2+74 x-6 x-111=0$
$\Rightarrow 4 x^2+74 x-6 x-111=0$
$\Rightarrow 2 x (2 x +37)-3(2 x +37)=0$
$\Rightarrow(2 x+37)(2 x-3)=0$
$\Rightarrow x =-\frac{37}{2} \text { or } x =\frac{3}{2}$
As width can't be negative.
$\Rightarrow x=\frac{3}{2}=1.5$
Therefore, the width of the path is $1.5 m.$
Length of the field including the path $= (20 + 2x) m$
Breadth of the field including the path $= (14 + 2x) m.$
Area of rectangle $= L B$
Area of the field including the path $= (20+2 x)(14+2 x) m ^2$.
Area of the field excluding the path $= (20 \times 14) m ^2=280 m^2$
$\therefore$ Area of the path $=(20+2 x)(14+2 x)-280$
$(20 + 2x) (14 + 2x) - 280 = 111$
$\Rightarrow 4 x^2+68 x-111=0$
Factorise the equation,
$\Rightarrow 4 x^2+74 x-6 x-111=0$
$\Rightarrow 4 x^2+74 x-6 x-111=0$
$\Rightarrow 2 x (2 x +37)-3(2 x +37)=0$
$\Rightarrow(2 x+37)(2 x-3)=0$
$\Rightarrow x =-\frac{37}{2} \text { or } x =\frac{3}{2}$
As width can't be negative.
$\Rightarrow x=\frac{3}{2}=1.5$
Therefore, the width of the path is $1.5 m.$
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