We have $n$ isosceles-right-angled triangles. The hypotenuse of the $n^{\textrm{th}}$ triangle is the base of the $(n+1)^{\textrm{th}}$ triangle. For the first triangle, $T_{1}$, the length of the base is $1$ unit. What is the length of the hypotenuse of $T_{25}$?

One way of doing this, is to find the hypotenuses of all triangles (one-by-one). So the hypotenuse of the first triangle, $H_{1}=\sqrt{1^2+1^2}=\sqrt{2}$. The hypotenuse of the second triangle, $H_{2}=\sqrt{(\sqrt{2})^2+(\sqrt{2})^2}=2$, and so on until we reach $H_{25}$ which is the hypotenuse of $T_{25}$.

The mentioned way is tedious, is there a better way?

  • $\begingroup$ You might want to check your calculation for $H_2$. $\endgroup$ – Angina Seng Dec 4 '18 at 7:22
  • $\begingroup$ @LordSharktheUnknown you are right, thanks. $\endgroup$ – Hussain-Alqatari Dec 4 '18 at 7:27

We have $H_{n+1}= \sqrt{2}H_n$ from trigonometry.

That is $H_n$ is a geometric sequence with common ratio $\sqrt2$.

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  • $\begingroup$ Many thanks! Suppose that the hypotenuse of the $n^{\textrm{th}}$ triangle is the base of the $(n+1)^{\textrm{th}}$ triangle, and each triangle has a height of $1$ unit. If the base of the first triangle is $1$ unit, what is the hypotenuse of $T_{25}$? $\endgroup$ – Hussain-Alqatari Dec 4 '18 at 7:40
  • $\begingroup$ write out a few terms. there is a pattern. $\endgroup$ – Siong Thye Goh Dec 4 '18 at 7:41

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