The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 3

For Competitive Exams

Question 6 The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.

Solution:

Let us say the shorter side of the rectangle is x m.

Then, larger side of the rectangle = (x + 30) m

Diagonal of the rectangle =$\sqrt{x^{2}+(x+30)^{2}}$

As given, the length of the diagonal is = x + 30 m

Therefore,

$\sqrt{x^{2}+(x+30)^{2}}$=x+60

⇒ $x^{2}2 + (x + 30)^{2}$ = $(x + 60)^{2}$

⇒ $x^{2}+ x^{2} + 900 + 60x $= $x^{2} + 3600 + 120x$

⇒$ x^{2} – 60x – 2700$ = 0

⇒ $x^{2} – 90x + 30x – 2700$ = 0

⇒ x(x – 90) + 30(x -90) = 0

⇒ (x – 90)(x + 30) = 0

⇒ x = 90, -30

However, the side of the field cannot be negative. Therefore, the length of the shorter side will be 90 m, and the length of the larger side will be (90 + 30) m = 120 m.

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