A spiral of right-angled triangles
A spiral of right-angled triangles
IP 2
8 marks
Triangle \(1\) is right-angled with both shorter sides of length \(1\) unit. For \(n\ge2\), triangle \(n\) is right-angled; its shorter sides are the hypotenuse of triangle \(n-1\) and a side of length \(1\) unit. Let \(h_n\) be the length of the hypotenuse of triangle \(n\).
- Find \(h_1\), \(h_2\) and \(h_3\). Show that \(h_n=\sqrt{n+1}\) for every positive integer \(n\).[4]
- Find the first triangle whose hypotenuse is longer than \(5\) units.[2]
- Explain why exactly three of triangles \(1\) to \(15\) have a hypotenuse of whole-number length.[2]
Solution:
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Answer:(a) \(\sqrt{2}\), \(\sqrt{3}\), \(2\); \(h_n^2=h_{n-1}^2+1\) with \(h_1^2=2\) (b) Triangle \(25\) (c) \(n+1\) must be a perfect square: \(n=3\), \(8\), \(15\)
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