The diagonal of rectangular field is 60 meters more

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

Ans: Let’s consider the shorter side of the field be X m.

Hence, the longer side will be (X + 30) m (by given condition) …… (i)

Also the diagonal of the field will be (X + 60) m (by given condition) …… (ii)

Now, we know that the diagonal of a rectangle is given by: $D = \sqrt {(Side_1)^2 + (Side_2)^2}$

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

$\therefore (X + 60)^2 = (X^2 + (X + 30)^2)$

$\therefore (X^2 + 120 X + 3600) = (X^2 + (X^2 + 60 X + 900))$

$\therefore (X^2 + 120 X + 3600) = (2 X^2 + 60 X + 900)$

$\therefore X^2 - 60 X - 2700 = 0$

$\therefore X^2 - 90 X + 30 X - 2700 = 0$

$\therefore (X - 90) (X + 30) = 0$

Therefore, X = 90 and X = -30

Here, we reject X = – 30 because a side can not be negative value, and we accept X = 90 only

Hence, shorter side, X = 90 m

From Equation (ii), we get longer side = X + 30 = 90 + 30 = 120 m

Hence, sides of the squares are 90 m and 120 m.

Check:

If shorter sides of the filed is 90 m, and the longer side is 120 m, the diagonal of the field will be $\sqrt {90^2 + 120^2} = \sqrt {8100 + 14400} = \sqrt {22500} = 150 m$