Square root spiral

In a square root spiral, a unit segment is drawn perpendicular to the previous hypotenuse. By Pythagoras, the square of the new hypotenuse equals the previous square plus 1. A parallel segment would not form a right angle. Exam tip: remember “perpendicular unit segment.”

The two perpendicular sides of the first right triangle are 1 unit each. By Pythagoras’ theorem, the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). \(2\) is the value of the square of the hypotenuse, not the hypotenuse itself. Exam tip: the first hypotenuse in a square root spiral is always \(\sqrt{2}\).

The governing concept is the construction of a square-root spiral using the Pythagorean theorem. At this stage, the existing hypotenuse of length √2 becomes one leg of a new right triangle, and the newly drawn perpendicular is the other leg of length 1. If H is the next hypotenuse, then H² = (√2)² + 1² = 2 + 1 = 3. Since a length is positive, H = √3. Therefore option B is correct. Option A is only the previous hypotenuse. Option C would incorrectly add 4 or treat the lengths without squaring, while option D would add an extra unit to the squared result. The theorem must be applied to the squares of the perpendicular sides.

With (\sqrt{3}) and (1) unit, the new hypotenuse is (\sqrt{3+1}=\sqrt{4}). The next square root appears in order.

At every step, a (90^\circ) angle is made. Therefore each new triangle is a right triangle.

In a square root spiral, each new right triangle uses the hypotenuse of the previous triangle as one side, and its new perpendicular side is kept 1 unit long. By Pythagoras’ theorem, the successive hypotenuses become \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. \(\sqrt{2}\) is not the new perpendicular side; it is the hypotenuse of the first new triangle. Exam tip: remember that the side kept constant at every step is 1 unit.

Pythagoras theorem is used to find the hypotenuse. Recognizing the right triangle is important.

Adding a perpendicular 1-unit side forms a right triangle. By Pythagoras, the square of the new hypotenuse equals the previous square plus 1, producing successive square roots. Exam tip: identify the perpendicular unit side first.

The sequence of hypotenuses is (\sqrt{1},\sqrt{2},\sqrt{3}). Therefore the third hypotenuse is (\sqrt{3}).

Since \(2^2=4\) and \(3^2=9\), and 7 lies between 4 and 9, \(\sqrt{7}\) lies between 2 and 3. Do not choose 1 and 2: their squares bracket numbers only up to 4. Exam tip: compare with nearby perfect squares.

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

In a square root spiral, the previous hypotenuse becomes a side of the next triangle, and a new perpendicular side of 1 unit is added. The new hypotenuse follows from Pythagoras’ theorem. Exam tip: identify the newly added unit side at every step.

The new hypotenuse is (\sqrt{8+1}=\sqrt{9}). The next step always gives the square root of the next integer.

In a square root spiral, each new triangle must be a right-angled triangle. Therefore, the 1-unit perpendicular segment is drawn at \(90^\circ\) to the previous side. A line at \(60^\circ\) or \(45^\circ\) would not be perpendicular and would not form the required right triangle. Exam tip: the word “perpendicular” always indicates an angle of \(90^\circ\).

Adding (1) unit to (\sqrt{9}) gives the new hypotenuse (\sqrt{10}). Identify the previous square root.

At each step, the previous hypotenuse becomes one side and (1) unit perpendicular is the other side. The new hypotenuse gives the next square root.

The hypotenuse of length (\sqrt{2}) is taken in the compass and an arc is drawn on the number line. This locates the point.

The sequence is (\sqrt{1},\sqrt{2},\sqrt{3},\sqrt{4}). Therefore the fourth hypotenuse is (\sqrt{4}).

In a square root spiral, each new right triangle uses the previous hypotenuse and a unit side. \((\sqrt{12})^2+1^2=12+1=13\), so its hypotenuse is \(\sqrt{13}\). Exam tip: apply Pythagoras’ theorem.

In a square root spiral, each new right triangle uses the hypotenuse of the previous triangle as one of its sides, while the other side is 1 unit. By Pythagoras, the new hypotenuse becomes \(\sqrt{2},\sqrt{3},\sqrt{4}\), and so on. Exam tip: the shared side is always the previous hypotenuse.