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.

Direct answer: Option B, \(\sqrt{10}\). In the square-root spiral, the new right triangle has hypotenuse \(\sqrt{11}\) and one perpendicular side 1. Let the previous radius be \(r\). By Pythagoras, the square of the hypotenuse equals the sum of the squares of the perpendicular sides: \((\sqrt{11})^2=r^2+1^2\). Thus \(11=r^2+1\), so \(r^2=10\), and because a radius is positive, \(r=\sqrt{10}\). Option A, \(\sqrt{9}=3\), would give a hypotenuse squared of \(9+1=10\), not 11. Option B works exactly. Option C, 10, is the value of the radius squared, not the radius itself. Option D, \(\sqrt{12}\), would give a radius squared of 12 and is too large. The key is to subtract the known side’s square from the hypotenuse’s square, then take the positive square root. Memory cue: reverse Pythagoras means hypotenuse square minus known side square.

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.

(\sqrt{2}) is irrational, so its decimal value does not terminate. The spiral represents it exactly geometrically.

The governing idea of a square-root spiral is the Pythagorean theorem. If the existing hypotenuse has length √3 and a new perpendicular segment of length 1 is attached to its endpoint, the two perpendicular sides of the new right triangle are √3 and 1. Therefore, the square of the new hypotenuse is (√3)² + 1² = 3 + 1 = 4. Taking the positive square root, because a length is positive, gives √4, which is equal to 2. Hence option C is correct. Option A would result from adding only 2 rather than 1 to the squared length, option B repeats the old hypotenuse without incorporating the new segment, and option D incorrectly adds 2 to 3. The construction therefore represents the next square-root length, √4.

From (\sqrt{k+1}=\sqrt{21}), we get (k+1=21), so (k=20). In such questions, equate the numbers under the roots.

Then the hypotenuse is (\sqrt{2^2+1^2}=\sqrt{5}), so the standard spiral is not formed. It is necessary to start with (OA=1).

Each new right triangle has the previous hypotenuse as one leg and 1 as the other. By Pythagoras, new square = previous square + 1. Option B is wrong because the length itself does not rise by 1. Exam tip: compare squares of hypotenuses, not their lengths.