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Problem 6

A right-angled triangle has a hypotenuse of length $2 x - 3.$ The other two sides have lengths of $x$ and $x - 1.$

Right angled triangle

Find the value of $x .$

Hint

By Pythagoras' Theorem,

x^{2} + (x - 1)^{2} = (2 x - 3)^{2}

Solution

By Pythagoras' Theorem,

\begin{aligned} x^{2} + (x - 1)^{2} & = (2 x - 3)^{2} \\ x^{2} + x^{2} - 2 x + 1 & = 4 x^{2} - 12 x + 9 \\ 2 x^{2} - 10 x + 8 & = 0 \\ x^{2} - 5 x + 4 & = 0 \\ (x - 1) (x - 4) & = 0 \\ x & \in {1, 4} \end{aligned}

However, $x = 1$ does not make sense as a solution, since then the side of length $x - 1$ would be $0$ and the side of length $2 x - 3$ would be $- 1$ , neither of which make sense.

Therefore, $x = 4$ is the only solution.

Problem 6 ​

Problem 6