7. Teodoro is building a sequence of right triangles, arranged as shown on the side. The first is the isosceles triangle \(A_{1} A_{2} A_{3}\), right-angled at \(A_{1}\), with legs of \(1 \, \text{cm}\). The second is \(A_{2} A_{3} A_{4}\), right-angled at \(A_{2}\), where \(A_{2} A_{4}\) is still \(1 \, \text{cm}\). The third is \(A_{3} A_{4} A_{5}\), right-angled at \(A_{3}\), with \(A_{3} A_{5}\) still \(1 \, \text{cm}\). The construction continues in this way: in each triangle \(A_{n} A_{n+1} A_{n+2}\), right-angled at \(A_{n}\), the leg \(A_{n} A_{n+2}\) is always \(1 \, \text{cm}\). How many \(\text{cm}\) will the segment \(A_{100} A_{101}\) measure?
(A) 20
(B) 50
(C) 25
(D) 10
(E) 16

![](https://cdn.mathpix.com/cropped/2024_04_17_94b06074249d24c20b4bg-1.jpg?height=369&width=340&top_left_y=994&top_left_x=2407)
(7) The correct answer is $(D)$.

By the Pythagorean theorem, we have $A_{2} A_{3}=\sqrt{1^{2}+1^{2}}=\sqrt{2}$, so $A_{3} A_{4}=\sqrt{1^{2}+2}=\sqrt{3}$, and so on: $A_{n} A_{n+1}=\sqrt{1^{2}+(n-1)}=\sqrt{n}$. Therefore, $A_{100} A_{101}=10$.

Question proposed by Paolo Francini.