#### 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`

⇒ x^{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)

⇒ (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

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

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

2x^{2} + 60x + 900 - x^{2} - 120x - 3600 = 0

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 = 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.