The hypotenuse of

Let one side of right – angled triangle be x m and other side be x + 4 m

On applying the Pythagoras theorem –

20² = (x + 4)² + x²

400 = x² + 8x + 16 + x²

400 = 2x² + 8x + 16

2x² + 8x – 384 = 0

x² + 4x – 192 = 0

Using the splitting middle term – the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1 b = 4 c = – 192

= 1. – 192 = – 192

And either of their sum or difference = b

= 4

Thus the two terms are 16 and – 12

Difference = 16 – 12 = 4

Product = 16. – 12 = – 192

x² + 16x – 12x – 192 = 0

x(x + 16) – 12(x + 16) = 0

(x + 16) (x – 12) = 0

x = 16 or x = 12

x cannot be negative

Base is 12m and other side is 12 + 4 = 16m