71

The hypotenuse of a right – angled triangle is 1 metre less than twice the shortest side. If the third side is 1 metre more than the shortest side, find the sides of the triangle.

Let the shortest side of triangle be xm

According to the question ;

Hypotenuse = 2x – 1 m

Third side = x + 1 m

Applying Pythagoras theorem

(2x – 1)^{2} = (x + 1)^{2} + x^{2}

4x^{2} – 4x + 1 = x^{2} + 2x + 1 + x^{2} using (a – b)^{2} = a^{2} – 2ab + b^{2}

2 x^{2} – 6x = 0

2x (x – 3) = 0

x = 0 or x = 3

Length of side cannot be 0 thus the shortest side is 3m

Hypotenuse = 2x – 1 = 6 – 1 = 5m

Third side = x + 1 = 3 + 1 = 4m

Thus the dimensions of triangle are 3m, 4m and 5m.

71