Question

# A right-angle triangle has a hypotenuse of length $Q\mathrm{cm}$ and one side of length $P\mathrm{cm}$. If $Q-P=2$ express the length of the third side of the right-angle triangle in terms of $Q$.

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Solution

## Express the third side in terms of $Q$.According to the Pythagoras theorem, In a right-angle triangle, the square of the hypotenuse is equal to the sum of the square of the other two sides.Here, $\mathrm{P}\mathrm{cm}$ is the length of the hypotenuse and $Q$ is one of the other sides of right angle triangle.Assume $x$ the the third side of the triangle. Find the length of the third side of the right-angle triangle in terms of $Q$: ${\mathrm{Q}}^{2}={\mathrm{P}}^{2}+{\mathrm{x}}^{2}\left[\mathrm{by}\mathrm{pythagoras}\mathrm{theorem}\right]\phantom{\rule{0ex}{0ex}}⇒{x}^{2}={\mathrm{Q}}^{2}-{\mathrm{P}}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=\left(\mathrm{Q}-\mathrm{P}\right)\left(\mathrm{Q}+\mathrm{P}\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=2\left(\mathrm{Q}+\mathrm{P}\right)\mathbf{\left[}\mathbf{\because }\mathbf{}\mathbit{Q}\mathbf{-}\mathbit{P}\mathbf{=}\mathbf{2}\mathit{\right]}\phantom{\rule{0ex}{0ex}}\mathit{⇒}\mathit{}\mathit{}\mathit{}\mathit{}{x}^{\mathit{2}}\mathit{=}2\left(\mathrm{Q}+\mathrm{Q}-2\right)\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=4\mathrm{Q}-4\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=4\left(\mathrm{Q}-1\right)\phantom{\rule{0ex}{0ex}}⇒x=\sqrt{4\left(\mathrm{Q}-1\right)}\phantom{\rule{0ex}{0ex}}⇒x=2\sqrt{\mathrm{Q}-1}$ Hence, the third side length is $2\sqrt{Q-1}$.

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