Square root spiral

( \sqrt{16}=4 ), so it lies at integer distance (4). Points for perfect squares give integer distances in the spiral.

The spiral constructs hypotenuse lengths ( \sqrt{n} ). When (n) is not a perfect square, such lengths are irrational.

By the Pythagorean theorem, the new hypotenuse is \(\sqrt{(\sqrt{24})^2+1^2}=\sqrt{24+1}=\sqrt{25}=5\). Thus, it is not merely closest to 5; it is exactly 5. Although 4 and 6 are neighbouring integers, they cannot be correct because the new hypotenuse is \(\sqrt{25}\). Exam tip: in a square-root spiral, adding a unit perpendicular increases the square of the hypotenuse by 1.

The next hypotenuse is \( \sqrt{(\sqrt{6})^2+1^2}=\sqrt{7} \). Squares are added, not lengths.

The old hypotenuse ( \sqrt{2} ) makes a right angle with the new side (1). The new hypotenuse is ( \sqrt{3} ).

At each step, the number under the square root increases by (1). Therefore, after ( \sqrt{30} ), ( \sqrt{31} ) is formed.

If hypotenuses are counted from the first right triangle, the sequence starts at ( \sqrt{2} ). The base ( \sqrt{1} ) is the initial radius.

( \sqrt{50}=\sqrt{25\times 2}=5\sqrt{2} ). Extract the largest perfect square while simplifying.

The new hypotenuse is ( \sqrt{36}=6 ), exactly (6). Among the given options, it is associated with the (6) and (7) range.

For ( \sqrt{15} ), the previous radius is ( \sqrt{14} ) and the new side is (1). This sequence identifies the spiral.

(OP_{n+3}^2=n+3) and (OP_n^2=n), so the difference is (3). Working with squares makes the question easier.

Write the surd as a single square root: \(2\sqrt{3}=\sqrt{2^2\times3}=\sqrt{12}\). Therefore, in the square root spiral, the new hypotenuse occurs at the \(\sqrt{12}\) step. Note that \(\sqrt{9}=3\), so it is not equal to \(2\sqrt{3}\). Exam tip: use \(a\sqrt{b}=\sqrt{a^2b}\) to move a coefficient inside a square root.

Since \(40=4\times10\), and \(4\) is a perfect square, \(\sqrt{40}=\sqrt{4\times10}=\sqrt4\times\sqrt{10}=2\sqrt{10}\). The option \(\sqrt{20}\) is not in simplest form because \(\sqrt{20}=2\sqrt5\). Exam tip: first identify the greatest perfect-square factor while simplifying a surd.

Since (36=6^2) and (49=7^2), (6<\sqrt{48}<7). Comparing with perfect squares is always reliable.

Because ( \sqrt{4}=2 ), the distance (OP_4) is (2). Understand the index and the distance separately.

Since \(63=9\times 7\), and \(9\) is a perfect square, \(\sqrt{63}=\sqrt{9\times7}=\sqrt9\times\sqrt7=3\sqrt7\). \(7\sqrt3\) is not correct because its square is \(147\), not \(63\). Exam tip: To simplify a square root, first identify the greatest perfect-square factor of the number.

The square of the radius \(\sqrt{5}\) is \(5\), and the square of the radius \(\sqrt{13}\) is \(13\). Therefore, the difference is \(13-5=8\). \(\sqrt{8}\) is the square root of the difference, not the difference itself. Exam tip: use \((\sqrt{n})^2=n\) directly.

\( \sqrt{72}=\sqrt{36\times 2}=6\sqrt{2} \), so \(a=6\). Look for a perfect-square factor.

The square-root spiral uses the Pythagorean theorem at every stage. Suppose the current radius or hypotenuse has length \(\sqrt{n}\). A new perpendicular side of length 1 unit is drawn. Since the two sides are perpendicular, the new hypotenuse has square equal to the sum of the squares of the two sides: \((\sqrt{n})^2+1^2=n+1\). Taking the positive square root gives the new length \(\sqrt{n+1}\).

Option A is incorrect because \(\sqrt{n}+1\) is generally not equal to \(\sqrt{n+1}\); for example, when \(n=4\), the two values are 3 and \(\sqrt{5}\). Option B is also false because consecutive integers are not always perfect squares. Option C gives the wrong length for the new side: it is 1 unit, not \(n\) units. Therefore option D correctly explains the construction.

Since \(64<80<81\), \(8<\sqrt{80}<9\). Remembering perfect squares helps in such questions.