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?
Best Answer
We have $H_{n+1}= \sqrt{2}H_n$ from trigonometry.
That is $H_n$ is a geometric sequence with common ratio $\sqrt2$.