Square root spiral

By Pythagoras theorem, (OB^2=1^2+1^2=2), so (OB=\sqrt{2}). In exams, identify the right triangle first.

Here (OC^2=(\sqrt{2})^2+1^2=3), so (OC=\sqrt{3}). At each new step, (1) is added to the previous square under the root.

On (OC=\sqrt{3}), drawing a perpendicular of (1) unit gives (OD^2=3+1=4). Remembering the order is very useful in such questions.

The square of the new radius is (8+1=9), so the length is (\sqrt{9}). Remember that the length is written as (\sqrt{9}), not just (9).

Because ((\sqrt{6})^2+1^2=7), the new segment becomes (\sqrt{7}). Identifying the correct previous segment is necessary.

Putting (n=12) in the formula gives (OP_{12}=\sqrt{13}). Pay attention to the difference between the index and the number under the root.

From (\sqrt{2}) to (\sqrt{10}), the number under the root increases by (8), so (8) new steps are needed. Each step increases the inner number by (1).

In every new right triangle, a perpendicular side of (1) unit is added. This is why the next hypotenuse represents the next square root.

The square of the previous radius is (11-1=10), so it was (\sqrt{10}). In reverse questions, subtract (1).

In the square root spiral, the length itself does not increase by (1); its square increases by (1). This difference helps solve many difficult questions.

Since \(PQ \perp OP\), \(\triangle OPQ\) is right-angled and \(OQ\) is the hypotenuse. By the Pythagorean theorem, \(OQ^2=OP^2+PQ^2=(\sqrt{15})^2+1^2=15+1=16\). The option \(\sqrt{16}\) represents \(OQ=4\), whereas the question asks for \(OQ^2\). Exam tip: Check carefully whether the question asks for a length or its square.

The order is (\sqrt{2},\sqrt{3},\sqrt{4},\sqrt{5}). Therefore, (\sqrt{5}) comes immediately after (\sqrt{4}).

The whole construction of the spiral is based on right triangles. Pythagoras theorem gives the length of the new hypotenuse.

The first triangle gives (\sqrt{2}), and the fifth triangle gives (\sqrt{6}). So, (5) right triangles are formed.

The first triangle gives (\sqrt{2}), the second (\sqrt{3}), the third (\sqrt{4}), and the fourth (\sqrt{5}). Add (1) to the step number.

((\sqrt{12})^2+1^2=13), so the next perpendicular is drawn on the hypotenuse (\sqrt{12}). Identify the previous hypotenuse by reducing the inner number by (1).

In the next step, a perpendicular of (1) unit is added, so the new distance becomes (\sqrt{17+1}=\sqrt{18}). The number under the root increases.

For consecutive hypotenuses, the numbers under the roots are consecutive. Therefore, (\sqrt{9}) and (\sqrt{10}) form the correct pair.

In the immediately previous step, the number under the root is (1) less, so the length is (\sqrt{19}). Learn to read the sequence backward too.

(\sqrt{4}=2) and (\sqrt{9}=3), so both are rational. The spiral can show both rational and irrational square roots.