The World of Numbers | EOT

Question 16

Find the lengths of the hypotenuses of all the right triangles in Fig. 3.14 which is referred to as the square root spiral.

Solution

Understand the Question

The square root spiral (or Spiral of Theodorus) is constructed using consecutive right-angled triangles.
The first triangle starts with two perpendicular legs of unit length $1$ .
Each successive right triangle uses the hypotenuse of the previous triangle as its base and a perpendicular side of unit length $1$ .
By applying the Pythagoras theorem $(\text{Hypotenuse}^2 = \text{Base}^2 + \text{Perpendicular}^2)$ successively, the lengths of the hypotenuses form the sequence $\sqrt{2}, \sqrt{3}, \sqrt{4}, \sqrt{5}, \dots, \sqrt{n}$ .

Step 1 · Find the Hypotenuse of the First Triangle (OP₂)

In the first right-angled triangle $\Delta OP_1P_2$ :

$\text{Base } OP_1 = 1$
$\text{Perpendicular } P_1P_2 = 1$

Applying the Pythagoras theorem:

\begin{aligned} OP_2^2 &= OP_1^2 + P_1P_2^2 \\ &= 1^2 + 1^2 \\ &= 1 + 1 \\ &= 2 \\ OP_2 &= \sqrt{2} \end{aligned}

Step 2 · Find the Hypotenuse of the Second Triangle (OP₃)

In the second right-angled triangle $\Delta OP_2P_3$ :

$\text{Base } OP_2 = \sqrt{2}$
$\text{Perpendicular } P_2P_3 = 1$

Applying the Pythagoras theorem:

\begin{aligned} OP_3^2 &= OP_2^2 + P_2P_3^2 \\ &= (\sqrt{2})^2 + 1^2 \\ &= 2 + 1 \\ &= 3 \\ OP_3 &= \sqrt{3} \end{aligned}

Step 3 · Find the Hypotenuse of the Third Triangle (OP₄)

In the third right-angled triangle $\Delta OP_3P_4$ :

$\text{Base } OP_3 = \sqrt{3}$
$\text{Perpendicular } P_3P_4 = 1$

Applying the Pythagoras theorem:

\begin{aligned} OP_4^2 &= OP_3^2 + P_3P_4^2 \\ &= (\sqrt{3})^2 + 1^2 \\ &= 3 + 1 \\ &= 4 \\ OP_4 &= \sqrt{4} = 2 \end{aligned}

Step 4 · Generalise the Lengths for the Entire Spiral

Continuing this process for each subsequent right triangle $\Delta OP_{n-1}P_n$ , where the base is $OP_{n-1} = \sqrt{n-1}$ and the perpendicular is $P_{n-1}P_n = 1$ :

\begin{aligned} OP_n^2 &= OP_{n-1}^2 + P_{n-1}P_n^2 \\ &= (\sqrt{n-1})^2 + 1^2 \\ &= (n - 1) + 1 \\ &= n \\ OP_n &= \sqrt{n} \end{aligned}

Thus, the lengths of the hypotenuses of successive right triangles are: $OP_2 = \sqrt{2}, \quad OP_3 = \sqrt{3}, \quad OP_4 = \sqrt{4} = 2, \quad OP_5 = \sqrt{5}, \quad \dots, \quad OP_n = \sqrt{n}$

Answer

$\sqrt{2}, \sqrt{3}, \sqrt{4} = 2, \sqrt{5}, \dots, \sqrt{n}$

Common Mistakes

Squaring Radicals Incorrectly: Forgetting that $(\sqrt{k})^2 = k$ , which leads to incorrect algebraic sums under the square root.
Varying Perpendicular Length: Assuming the perpendicular leg changes length; in a square root spiral, every perpendicular leg is always fixed at $1\text{ unit}$ .

Question 16

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