# What is the length of the hypotenuse?

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?

• You might want to check your calculation for $H_2$. – Lord Shark the Unknown Dec 4 '18 at 7:22
• @LordSharktheUnknown you are right, thanks. – 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 $$\sqrt2$$.
• 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 $T_{25}$? – Hussain-Alqatari Dec 4 '18 at 7:40