Jeremy loves right triangles that have side length 1. In fact, he loves to combine them so that the hypotenuse of one becomes the leg of the next.

Question

If the longer leg of the first
triangle is sqrt 11, what will be the hypotenuse of the 2014th triangle?
I don't understand what I have to do to solve the problem.

Answer 1

Start with T1 which has sides 1 and 1. Its hypotenuse is √2

Now draw T2, with its base as the hypotenuse of T1, and its other side is 1.

T2 has sides √2 and 1, so its hypotenuse is √3

Keep drawing new triangles on the hypotenuse, and you will get a spiral figure.

Tn has its sides √n and 1, so its hypotenuse is √(n+1)

Answer 2

The answer is actually 45

Answer 3

We use the Pythagorean Theorem to find the first few hypotenuses:

[asy]
pair A=(0,0),B=(0,1),C=(0.4404,1.957), D=(1.2,2.9), E=(3.316,0);
draw(A--B--C--D--E--A);
draw(B--E--C);
draw(D--E);
draw(rightanglemark(E,A,B));
draw(rightanglemark(E,B,C));
draw(rightanglemark(E,C,D));
label("1",midpoint(A--B),W);
label("1",midpoint(C--B),NW);
label("1",midpoint(C--D),NW);
label("...",(2,2),E);
defaultpen(fontsize(8pt));
label("$\sqrt{11}$",midpoint(A--E),N);
label("$\sqrt{12}$",midpoint(B--E),N);
label("$\sqrt{13}$",midpoint(C--E),NE);
label("$\sqrt{14}$",midpoint(D--E),NE);
[/asy]

So, the first triangle has hypotenuse $\sqrt{12}$, the 2nd has hypotenuse $\sqrt{13}$, and so on. Thus, the 2014th will have hypotenuse $\sqrt{2014+11}=\sqrt{2025}=\boxed{45}$.