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Bunuel
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siddhantmittal
We know that area of rectangle is length x breadth

Approach 1:
Start with option (c).. we know that if width of the track is 4m (and since it is a uniform track)
Area of park \(= 14*10 = 140 m^2\)
New length \(= 14 + 4 + 4 = 22m\)
New breadth \(= 10 + 4 + 4 = 18m\)

Area of path \(= (22*18 ) - (140) m ^2\)

Since this will give us 6 in the units place, we can rule out this option

If we plug in the value of option (b), we see that the difference of area matches with the area of path given
Hence the answer

Approach 2:
Consider the width of the track as x
New length \(= 14 + 2x\)
New breadth \(= 10 + 2x\)

Area of path\( = 112 = (14 + 2x)*(10+ 2x) - 140\)
Solve the quadratic to obtain the answer

Why do we add 2x to the new length and width instead of x? As in 14 + x and 10 + x?
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CEdward
siddhantmittal
We know that area of rectangle is length x breadth

Approach 1:
Start with option (c).. we know that if width of the track is 4m (and since it is a uniform track)
Area of park \(= 14*10 = 140 m^2\)
New length \(= 14 + 4 + 4 = 22m\)
New breadth \(= 10 + 4 + 4 = 18m\)

Area of path \(= (22*18 ) - (140) m ^2\)

Since this will give us 6 in the units place, we can rule out this option

If we plug in the value of option (b), we see that the difference of area matches with the area of path given
Hence the answer

Approach 2:
Consider the width of the track as x
New length \(= 14 + 2x\)
New breadth \(= 10 + 2x\)

Area of path\( = 112 = (14 + 2x)*(10+ 2x) - 140\)
Solve the quadratic to obtain the answer

Why do we add 2x to the new length and width instead of x? As in 14 + x and 10 + x?

Because the uniform width x is on all sides, so length will increase by uniform width x increase in south direction and x increase in north direction. Same with width increase in east and west direction by x units.

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