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Question
The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Solution 1
Let the shorter side of the rectangle be x m.
Then, larger side of the rectangle = (x + 30) m
Diagonal of rectangle = `sqrt(x^2+(x+30)^2)`
It is given that the diagonal of the rectangle = (x+30)m
`:.sqrt(x^2+(x+30)^2) = x +60`
⇒ x2 + (x + 30)2 = (x + 60)2
⇒ x2 + x2 + 900 + 60x = x2 + 3600 + 120x
⇒ x2 - 60x - 2700 = 0
⇒ x2 - 90x + 30x - 2700 = 0
⇒ x(x - 90) + 30(x -90)
⇒ (x - 90)(x + 30) = 0
⇒ x = 90, -30
However, side cannot be negative. Therefore, the length of the shorter side will be 90 m.
Hence, length of the larger side will be (90 + 30) m = 120 m.
Solution 2
Let the length of smaller side of rectangle be x meters then larger side be (x + 30) meters and their diagonal be (x + 60)meters
Then, as we know that Pythagoras theorem
x2 + (x + 30)2 = (x + 60)2
x2 + x2 + 60x + 900 = x2 + 120x + 3600
2x2 + 60x + 900 - x2 - 120x - 3600 = 0
x2 - 60x - 2700 = 0
x2 - 90x + 30x - 2700 = 0
x(x - 90) + 30(x - 90) = 0
(x - 90)(x + 30) = 0
x - 90 = 0
x = 90
Or
x + 30 = 0
x = -30
But, the side of rectangle can never be negative.
Therefore, when x = 90 then
x + 30 = 90 + 30 = 120
Hence, length of smaller side of rectangle be 90 meters and larger side be 120 meters.
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