Question

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

Answer 1

Let the length of shortest side = x m
Length of hypotenuse = 2x – 1
and third side = x + 1
Now according to the condition,
(2x – 1)² = (x)² + (x + 1)²  ...(By Pythagorus Theorem)
⇒ 4x² – 4x + 1 = x² + x² + 2x + 1
⇒ 4x² – 4x + 1 = 2x² - 2x – 1 = 0
⇒ 2x² – 6x = 0
⇒ x² – 3x = 0
⇒ x(x – 3) = 0  ...(Dividing by 2)
Either x = 0,
but it is not possible
or
x – 3 = 0,
then x = 3
Shortest side = 3m
Hypotenuse = 2 x 3 – 1 = 6 – 1 – 5
Third side = x + 1 = 3 + 1 = 4
Hence sides are 3, 4, 5 (in m).