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.

Answer 1

Let shorter side of rectangle ABCD be x metre, Then, longer side of rectangle ABCD, 

AB = (x + 30) m. 

Diagonal AC = (x + 60) m. 

In ⊥∆ABC, ∠B = 90°. 

As per Pythagoras theorem, 

AB² + BC² = AC² 

(x + 30)² + (x)² = (x + 60)² 

x² + 60x + 900 + x² = x² + 120x + 3600 

2x² + 60x + 900 = x² + 120x + 3600 

2x² – x² + 60x – 120x + 900 – 3600 = 0

x² – 60x – 2700 = 0

x² – 90x + 30x – 2700 = 0 

x(x – 90) + 30 (x – 90) = 0 

(x – 90) (x + 30) = 0 

If x – 90 = 0, then x = 90 

If x + 30 = 0, then x = -30 

Shorter side of rectangle, BC = x = 90 m. 

Longer side of rectangle, AB = x + 30 = 90 + 30= 120 m. 

Diagonal of rectangle, AC = x + 60 = 90 + 60 = 150 m.