Square root spiral

A square root spiral is made from successive right triangles, each with a new perpendicular side of 1 unit. By Pythagoras, the hypotenuses become
1,
2,
3, and so on. Exam tip: identify the constant 1-unit added side.

To construct \(\sqrt{7}\), draw a perpendicular side of length 1 at the previous hypotenuse \(\sqrt{6}\). Since \((\sqrt{6})^2+1^2=7\), the new hypotenuse is \(\sqrt{7}\). \(\sqrt{5}\) belongs to an earlier step. Exam tip: add 1 at each new step.

Adding a (1) unit perpendicular to (\sqrt{10}) forms (\sqrt{11}). Remember the stepwise construction.

With a compass, the hypotenuse length is taken and an arc is drawn on the number line. This is the geometric construction.

By Pythagoras theorem, the new hypotenuse becomes (\sqrt{2+1}=\sqrt{3}). Do not treat (\sqrt{2}+1) as (\sqrt{3}).

With (\sqrt{3}) and (1) unit, the new hypotenuse is (\sqrt{4}). The correct calculation uses sum of squares.

Successive right triangles turn around and make a spiral-like shape. Hence it is called a square root spiral.

By Pythagoras, the square of the new hypotenuse is \((\sqrt{6})^2+1^2=6+1=7\). Hence it represents \(\sqrt{7}\). \(\sqrt{6}\) is the previous hypotenuse. Exam tip: square the old length first.

Adding a (1) unit perpendicular to (\sqrt{12}) forms (\sqrt{13}). The previous hypotenuse has one less number.

In a square root spiral, the previous hypotenuse becomes one side of the next right triangle, and the perpendicular new side is 1 unit. By Pythagoras’ theorem, the new hypotenuse is the square root of the next natural number. Exam tip: identify the added 1-unit side.

A square root spiral uses right triangles to create successive lengths. If the old hypotenuse is \(\sqrt{n}\) and a new perpendicular of length 1 is drawn at a right angle, the old hypotenuse becomes one leg of the larger right triangle. The new hypotenuse is therefore found from the squares of the two perpendicular sides, not by simply adding lengths.

By Pythagoras, the new length is \(\sqrt{(\sqrt{n})^2+1^2}\). Since \((\sqrt{n})^2=n\) and \(1^2=1\), this becomes \(\sqrt{n+1}\). Thus option A is correct. The other choices either subtract, double, or square the wrong quantity.

A square root spiral is constructed by repeatedly making a right triangle. At each stage, one existing hypotenuse is used as a side, and a new perpendicular side of length 1 unit is drawn. The Pythagorean theorem then gives the length of the new hypotenuse. If the old hypotenuse represents
sqrt{n}
, the new one represents
sqrt{n+1}
.

Thus, adding a 1-unit perpendicular is not meant to erase a line, change a denominator, or turn a triangle into a circle. It creates the next right triangle in the construction. Its hypotenuse has square length
(
sqrt{n}
)^2+1^2=n+1
, so its length is
sqrt{n+1}
. Therefore option A correctly states the purpose: to make the next square root appear as the hypotenuse.

In a square root spiral, the hypotenuse of one right triangle becomes a side of the next triangle, and a perpendicular side of length 1 is added. Thus, each new hypotenuse represents the next square root. Exam tip: track the previous hypotenuse at every step.

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

With previous hypotenuse \(\sqrt{n}\) and a new perpendicular side of 1 unit, Pythagoras gives \(\sqrt{n+1}\). A parallel side does not form the required right triangle. Exam tip: identify the fixed 1-unit leg.

The preceding hypotenuse becomes one leg, and a perpendicular unit leg is added at its endpoint. Pythagoras gives next hypotenuse² = previous hypotenuse² + 1. Exam tip: spot the unit perpendicular.

In a square root spiral, one leg of every new right triangle is fixed at 1 unit, while the other leg is the previous hypotenuse. For example, after \(\sqrt{5}\), the next hypotenuse is \(\sqrt{5+1}=\sqrt{6}\). Exam tip: remember that the fixed leg is always 1 unit.

With (\sqrt{17}) and a (1) unit perpendicular, (\sqrt{18}) is formed. The previous number is always (1) less.

In a square-root spiral, successive perpendicular unit segments create right triangles whose hypotenuse lengths are √2, √3, √4, and so on. Continuing the construction until the required stage produces a hypotenuse of length √20. That hypotenuse is transferred with the compass to the number line. Therefore option B is correct; √19 is the preceding stage and 20 units is not the required length.

Many hypotenuses like (\sqrt{2},\sqrt{3},\sqrt{5}) are not whole numbers. So this statement is incorrect.