Question

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

Answer 1

Let the shortest side be $x$ m.
Therefore, according to the question:
Hypotenuse = $\left(2x-1\right)$ m
Third side = $\left(x+1\right)$ m
On applying Pythagoras theorem, we get:

$$\begin{gathered} {(2x-1)}^2={(x+1)}^2+x^2 \\ \Rightarrow 4x^2-4x+1=x^2+2x+1+x^2 \\ \Rightarrow 2x^2-6x=0 \\ \Rightarrow 2x(x-3)=0 \\ \Rightarrow x=0\mathrm{or}x=3 \end{gathered}$$

The length of the side cannot be 0; therefore, the shortest side is 3 m.
Therefore,
Hypotenuse = $\left(2\times 3-1\right)=$5 m
Third side = (3 + 1) = 4 m