Square root spiral

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 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.

In the square root spiral, the next hypotenuse is formed from the previous hypotenuse and a perpendicular unit side. Here the two legs have lengths \\(\\sqrt{11}\\) and 1, so the new hypotenuse is not either individual leg. It is the segment opposite the right angle, represented by OQ in the construction, and its length is \\(\\sqrt{12}\\). Thus option D is correct.

Applying the theorem gives \\(h^2=(\\sqrt{11})^2+1^2=11+1=12\\). Taking the non-negative square root, \\(h=\\sqrt{12}\\). Although \\(\\sqrt{12}=2\\sqrt{3}\\) is another simplified form, the option uses the spiral’s step notation \\(\\sqrt{12}\\). Options A and B are legs, not the hypotenuse, while option C incorrectly removes the square root.

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.

( \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 direct answer is B: the next hypotenuse is sqrt{31}. In the standard square-root spiral, each new right triangle is formed by using the previous hypotenuse as one side and adding a perpendicular unit side. By the Pythagorean theorem, if the previous length is sqrt{30}, the next squared length is sqrt{30}^{2}+1^{2}=30+1=31. Therefore the new hypotenuse is sqrt{31}. Option A, sqrt{29}, goes backward by one and would represent the preceding construction, not the immediate next one. Option B, sqrt{31}, increases the number under the root by exactly one and is correct. Option C, 31, is the square of the new length, not the length itself; the hypotenuse is sqrt{31}, not 31. Option D, sqrt{60}, changes the number by 30 and does not follow the construction rule. The memory cue is: in this spiral, each step changes sqrt{n} into sqrt{n+1}, because one unit perpendicular side is added.

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

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.

Direct answer: Option C, \(3\). We are asked for the difference between the squares, not the difference between the lengths. Given \(OP_n=\sqrt n\), squaring gives \(OP_n^2=n\). Replacing \(n\) by \(n+3\), we get \(OP_{n+3}=\sqrt{n+3}\), so \(OP_{n+3}^2=n+3\). Now subtract: \(OP_{n+3}^2-OP_n^2=(n+3)-n=3\). Option A, 1, would represent an increase of only one in the index, not three. Option B, 2, similarly does not match the change from \(n\) to \(n+3\). Option C is correct. Option D, \(\sqrt3\), would be relevant to a difference of lengths in a different calculation, but it is not the difference of the squares requested here. Memory cue: square first, substitute second, subtract last.

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 \(63=9\times 7\), 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.

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 \(45=9\times5\) and \(9\) is a perfect square, \(\sqrt{45}=\sqrt{9\times5}=\sqrt9\times\sqrt5=3\sqrt5\). Therefore, option C is correct. \(5\sqrt3\) is not correct because its square is \(75\), not \(45\). Exam tip: simplify a surd by taking out its greatest perfect-square factor.

( \sqrt{99} ) is slightly less than (10) because (99) is close to (100). Compare with the perfect square (100).

The quantities \\(OP_a\\) and \\(OP_b\\) have lengths \\(\\sqrt{a}\\) and \\(\\sqrt{b}\\). Squaring them removes the square roots, giving \\(OP_a^2=a\\) and \\(OP_b^2=b\\). Therefore the requested difference is simply the difference between the indexed numbers. Since \\(b=a+5\\), the difference is 5, so option C is correct.

The calculation is \\(OP_b^2-OP_a^2=b-a=(a+5)-a=5\\). No square root remains after squaring the lengths. Option A incorrectly takes the square root of the difference, and option B doubles it. Option D gives the value of b rather than the difference. The result is independent of the particular value of a, provided the stated lengths are defined.

Integer distances occur at perfect squares, and the next perfect square after (28) is (36). So the next integer distance is at ( \sqrt{36}=6 ).