Square root spiral

By Pythagoras, the hypotenuse is ( \sqrt{1^2+1^2}=\sqrt{2} ). In exams, connect the first triangle with ( \sqrt{2} ).

The new radius is ( \sqrt{(\sqrt{8})^2+1^2}=\sqrt{9}=3 ). Remember that each new step adds (1) inside the square root.

For ( \sqrt{5} ), a unit segment (1) is drawn perpendicular to ( \sqrt{4} ). Perpendicularity is the key construction clue.

Each new right triangle uses an outer side of (1) unit. This gives (OP_{n+1}=\sqrt{n+1}).

A right angle is necessary for Pythagoras. Without a perpendicular side, ( \sqrt{n+1} ) is not guaranteed.

From ( \sqrt{2} ) to ( \sqrt{10} ), the inside number increases (8) times. Each increase needs one new unit segment.

The next triangle has perpendicular sides ( \sqrt{6} ) and (1). Its area is ( \frac{1}{2}\times \sqrt{6}\times 1=\frac{\sqrt{6}}{2} ).

Since (13) is not a perfect square, ( \sqrt{13} ) is irrational. The square root spiral shows such square roots geometrically.

Since \(OP_{k+1}=\sqrt{k+1}\), \(k+1=17\) and \(k=16\). Track the one-step increase carefully.

Because ( \sqrt{9}=3 ), the distance is (3) units. Simplify square roots of perfect squares immediately.

A square root spiral is built by repeatedly making a right triangle. At one stage, the previous distance from the centre is the hypotenuse of the existing construction, and a new side of length 1 is drawn at a right angle. If the old distance is \\(OP_n\\), the new hypotenuse is \\(OP_{n+1}\\).

By the Pythagorean theorem, the square of the new hypotenuse equals the sum of the squares of the two perpendicular sides. Therefore, \\(OP_{n+1}^2=OP_n^2+1^2=OP_n^2+1\\). This is exactly option C. Option A adds 1 to the length itself, which is not generally true; option B ignores the recursive construction, and D fixes every distance at 1.

The new length is ( \sqrt{5^2+1^2}=\sqrt{26} ). The length (1) is not added directly; squares are added.

The lengths of the two radii are \(\sqrt{20}\) and \(\sqrt{21}\). The difference between their squares is \((\sqrt{21})^2-(\sqrt{20})^2=21-20=1\). Hence, the correct answer is 1. \(\sqrt{21}-\sqrt{20}\) is the difference between the radii themselves, not the difference between their squares. Exam tip: squaring a square root directly gives its radicand.

The old hypotenuse is \(\sqrt{14}\), so its square is \(14\). Adding a perpendicular side of length 1 gives the square of the new hypotenuse as \(14+1^2=15\), so the new hypotenuse is \(\sqrt{15}\). Therefore, the ratio of the squares of the new and old hypotenuses is \(15:14\). Option D gives the ratio of the hypotenuses, not the ratio of their squares. Exam tip: In such questions, use the Pythagorean theorem to find the square of the hypotenuse first.

From ( \sqrt{1} ) to ( \sqrt{11} ), there are (11) points. Include the initial ( \sqrt{1} ) in counting.

( \sqrt{18} ) is formed by adding a perpendicular (1) to the previous hypotenuse ( \sqrt{17} ). Always check the previous square root in the sequence.

At each step, the new unit side is perpendicular to the previous radius. Thus Pythagoras gives \(r_{\text{new}}^2=r_{\text{old}}^2+1\), not \(r_{\text{old}}+1\). Exam tip: compare squares of successive radii.

The hypotenuse is ( \sqrt{(\sqrt{11})^2+1^2}=\sqrt{12} ). The step is identified by the new hypotenuse.

At every new step, the added side is (1) unit and perpendicular. The old radius keeps changing.

In a square-root spiral, the distance of the final point from the origin is the corresponding square root, here \(\sqrt{27}\). Since \(25<27<36\), with \(25=5^2\) and \(36=6^2\), we get \(5<\sqrt{27}<6\). Therefore, the distance lies between 5 and 6. The interval 4 to 5 is incorrect because its squared bounds run from \(16\) to \(25\). Exam tip: To locate a square root, compare the number with the nearest perfect squares on either side.