Square root spiral

A square root spiral is formed using successive right-angled triangles. The (90^\circ) angle allows the use of Pythagoras’ theorem to find each new hypotenuse, producing lengths such as \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. Equilateral triangles cannot produce this construction. Exam tip: If successive hypotenuses represent square roots, look for a right triangle and Pythagoras’ theorem.

The construction follows one repeated Pythagorean pattern. Suppose the current hypotenuse has length \(\sqrt{n}\), and a new perpendicular segment of length 1 is drawn from its endpoint. If the next hypotenuse is \(L\), then \(L^2=(\sqrt{n})^2+1^2=n+1\). Taking the positive square root, because a length is positive, gives \(L=\sqrt{n+1}\).

Thus the correct formula is option A. The expression \(\sqrt{n-1}\) would represent a smaller value, \(\sqrt{2n}\) would require a different construction, and \(\sqrt{n+2}\) would add 2 rather than the square of a unit side. The formula accurately describes every successive step of the spiral.

A square root spiral gives lengths like (\sqrt{2},\sqrt{3},\sqrt{5}). These can be placed on the number line.

The new right triangle has legs \(\sqrt{8}\) and 1. So its hypotenuse is \(\sqrt{(\sqrt{8})^2+1^2}=\sqrt{9}=3\). Adding \(\sqrt{8}+1\) is incorrect because a hypotenuse is not the ordinary sum of the legs. Exam tip: apply Pythagoras at every spiral step.

Drawing a (1) unit perpendicular on (\sqrt{11}) gives (\sqrt{12}). The previous hypotenuse has one less number.

At each step the next hypotenuse is (\sqrt{n+1}). Therefore (\sqrt{16}) is formed after (\sqrt{15}).

Since \(3^2=9\) and \(4^2=16\), any square root of a number between 9 and 16 lies between 3 and 4. Therefore, \(\sqrt{10}\) fits. \(\sqrt{17}\) is greater than 4. Exam tip: compare with nearby perfect squares first.

Both sides of the first triangle are (1) unit. So the hypotenuse is (\sqrt{1^2+1^2}=\sqrt{2}).

(\sqrt{2}) is the previous hypotenuse and (1) unit is the new perpendicular. Therefore the new hypotenuse is (\sqrt{3}).

In the usual square root spiral (1^2) is added each time. Taking (2) units will change the sequence.

By Pythagoras’ theorem, \(h_{new}^2=h_{old}^2+1^2\). Hence, each step adds 1 inside the square, producing successive square roots. The 1-unit segment is a leg, not the hypotenuse. Exam tip: identify the perpendicular legs before applying the theorem.

(\sqrt{21}) is formed from (\sqrt{20}) and a (1) unit perpendicular. The previous hypotenuse has one less number.

Each new right triangle uses a fresh side of length 1 unit and the previous hypotenuse as its other leg. By Pythagoras, the new hypotenuse becomes \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. Two unit legs would not describe later triangles. Exam tip: identify the previous hypotenuse first.

\(\sqrt{8}=2\sqrt{2}\), and \(\sqrt{2}\) is irrational; therefore, \(\sqrt{8}\) is irrational. In contrast, \(\sqrt{1}=1\), \(\sqrt{4}=2\), and \(\sqrt{9}=3\) are rational because 1, 4, and 9 are perfect squares. Exam tip: simplify the number under the root using a perfect-square factor, for example \(8=4\times2\), before deciding.

Each new triangle in a square root spiral is constructed by drawing a perpendicular side of 1 unit on the previous hypotenuse. By Pythagoras, the new hypotenuse becomes \(\sqrt{n+1}\). Using 2 units would not produce the standard spiral. Exam tip: remember the repeated added side is 1 unit.

A unit segment perpendicular to the previous hypotenuse forms the next right triangle. By Pythagoras,

ot valid?

A square root spiral is built by repeatedly adding a right triangle with one new perpendicular side of length 1. If the existing hypotenuse has length \(\sqrt{n}\), then the next hypotenuse has length \(\sqrt{n+1}\), because the Pythagorean theorem gives \((\sqrt{n})^2+1^2=n+1\). The construction therefore proceeds in order: \(\sqrt{28}\), \(\sqrt{29}\), \(\sqrt{30}\), and so on.

Just before constructing \(\sqrt{30}\), the existing hypotenuse must be \(\sqrt{29}\). A perpendicular segment of length 1 is then drawn, giving \((\sqrt{29})^2+1^2=29+1=30\), so the new hypotenuse is \(\sqrt{30}\). Hence option B is correct. \(\sqrt{28}\) is one stage earlier, while \(\sqrt{30}\) is the hypotenuse being made, not the one already present.

Using a compass the hypotenuse length is transferred to the number line by drawing an arc. This gives the square root point.

In a square root spiral the new hypotenuse gives the next square root in order. It is made from right triangles.

In a square root spiral, each new right triangle uses the previous hypotenuse and a perpendicular side of 1 unit. After \(\sqrt{9}\), the hypotenuse is \(\sqrt{9+1}=\sqrt{10}\). Using \(\sqrt{9}\) as the new side is incorrect. Exam tip: the added side is always 1 unit.