Square root spiral

Adding a (1) unit perpendicular to (\sqrt{17}) gives (\sqrt{18}). The previous hypotenuse has one less number.

In a square root spiral, one leg of every new right triangle is 1 unit, while the other leg is the previous hypotenuse. By Pythagoras’ theorem, the hypotenuses become \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. Exam tip: remember that the fixed added side is always 1 unit.

By Pythagoras’ theorem, the square of the new hypotenuse is \((\sqrt{n})^2+1^2=n+1\), so its length is \(\sqrt{n+1}\). The triangles need not be isosceles. Exam tip: follow the sequence of squared hypotenuse lengths.

If the previous hypotenuse is \(\sqrt{n}\) and a unit segment is drawn perpendicular to it, the square of the new hypotenuse is \(n+1\). Hence its length is \(\sqrt{n+1}\). In exams, first identify the right angle.

At each step of a square root spiral, a new side of length 1 unit is drawn at 90° to the previous side, forming a right-angled triangle. Pythagoras’ theorem then gives successive hypotenuse lengths such as \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. Option B is incorrect because the new hypotenuse is generally longer than the previous side. Exam tip: A 90° angle in such a construction signals the use of a right triangle and Pythagoras’ theorem.

Adding a (1) unit perpendicular to (\sqrt{49}) gives (\sqrt{50}). The previous number is (1) less.

In a square root spiral, each new right triangle has one leg of length 1, so its hypotenuse becomes \(\sqrt{2}, \sqrt{3}\), and so on. Thus, the distance from the starting point to the relevant point represents \(\sqrt{n}\). Exam tip: identify the hypotenuse, not the unit side.

In a square root spiral, the previous hypotenuse becomes one side and a perpendicular side of 1 unit is added. The new hypotenuse then represents √2, √3 and so on. Exam tip: remember that the added perpendicular side is always 1 unit.

Adding a (1) unit perpendicular to (\sqrt{6}) forms (\sqrt{7}). This follows from Pythagoras theorem.

Each new right triangle in a square root spiral adds a unit side. If the previous hypotenuse is \(\sqrt{4}\), Pythagoras gives the next hypotenuse as \(\sqrt{4+1}=\sqrt{5}\). \(\pi\) is not obtained in this sequence. Exam tip: remember the hypotenuse pattern \(\sqrt{2},\sqrt{3},\sqrt{4}\dots\).

In a square root spiral, a perpendicular side of length 1 unit is added to the previous hypotenuse. Thus, the new hypotenuses represent \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. Exam tip: the added side is always 1 unit.

With (\sqrt{14}) and a (1) unit perpendicular, (\sqrt{15}) is formed. The previous hypotenuse has one less number.

In a square root spiral, each new right triangle uses the previous hypotenuse and a new perpendicular side of length 1. Hence the hypotenuses increase, not remain equal. Exam tip: remember \(h_{\text{new}}^2=h_{\text{old}}^2+1\).

The hypotenuse length of the required square root is taken in the compass. So the hypotenuse of (\sqrt{3}) is needed.

In a square root spiral, a unit perpendicular is added to the previous hypotenuse. By Pythagoras, the new hypotenuse has square

a^2+1

; hence successive roots are formed. A parallel segment would not ensure a right triangle. Exam tip: look for “unit perpendicular.”

While constructing a square root spiral, each new triangle is formed by drawing a right angle on the previous hypotenuse. A set square helps construct this right angle accurately, so option A is correct. The length of a hypotenuse is generally measured with a ruler; the main use of a set square is constructing angles. Exam tip: at each step of a square root spiral, draw a new unit-length side perpendicular to the previous hypotenuse.

At each step, a unit segment is drawn perpendicular to the previous hypotenuse at its endpoint. If the previous hypotenuse is \(\sqrt{n}\), Pythagoras gives the next one as \(\sqrt{n+1}\). In exams, check the right-angle mark.

Adding a (1) unit perpendicular to (\sqrt{79}) gives (\sqrt{80}). The previous hypotenuse has one less number.

By Pythagoras’ theorem, new hypotenuse² = \((\sqrt{10})^2+1^2=10+1=11\). Hence the new hypotenuse is \(\sqrt{11}\). \(\sqrt{20}\) results from squaring the previous hypotenuse incorrectly. Exam tip: square the radical first.

A (1) unit perpendicular is drawn at the end of (\sqrt{2}) to make the next triangle. This gives (\sqrt{3}).