#### Question

The diagonal of a rectangle is 60 m more than its shorter side and the larger side is 30 m more than the shorter side. Find the sides of the rectangle.

#### Solution

Let the shorter side be x m.

Length of the other side = (x + 30) m

Length of hypotenuse = (x + 60) m

Using Pythagoras theorem,

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

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

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

But, x cannot be negative. So, x = 90.

Thus, the sides of the rectangle are 90 m and (90 + 30) m = 120 m.

Is there an error in this question or solution?

Solution The Diagonal of a Rectangle is 60 M More than Its Shorter Side and the Larger Side is 30 M More than the Shorter Side. Find the Sides of the Rectangle. Concept: Quadratic Equations.