Question

The hypotenuse of a right triangle is 6 m more than twice of the shortest side. If the third side is 2 m less than the hypotenuse, find the sides of the triangle

Answer 1

Let the shortest side of the right triangle be x.

∴ Hypotenuse = 6 + 2x

Third side = 2x + 6 – 2

= 2x + 4

In the right triangle ABC,

AC² = AB² + BC²

(2x + 6)² = x² + (2x + 4)²

4x² + 36 + 24x = x² + 4x² + 16 + 16x

0 = x² – 24x + 16x – 36 + 16

∴ x² – 8x – 20 = 0

(x – 10) (x + 2) = 0

x – 10 = 0 or x + 2 = 0

x = 10 or x = – 2 ...(Negative value will be omitted)

The side AB = 10 m

The side BC = 2(10) + 4 = 24 m

Hypotenuse AC = 2(10) + 6 = 26 m