Square root spiral

Correct answer: C. (OP_{n+1}^2=OP_n^2+1)

Explanation: 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.

Correct answer: A. 1

Explanation: 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.

Correct answer: C. \(15:14\)

Explanation: 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.

Correct answer: B. ( \sqrt{11} )

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

Correct answer: A. ( \sqrt{17} )

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

Correct answer: A. The statement is incorrect; the square of the new radius is 1 more than the square of the previous radius.

Explanation: 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.

Correct answer: D. ( \sqrt{12} )

Explanation: 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.

Correct answer: B. Between 5 and 6

Explanation: 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.

Correct answer: C. (4)

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

Correct answer: C. Because it geometrically gives distances like ( \sqrt{n} )

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

Correct answer: B. 5

Explanation: 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.

Correct answer: B. ( \sqrt{31} )

Explanation: 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.

Correct answer: C. ( \sqrt{2},\sqrt{3},\sqrt{4},\sqrt{5} )

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

Correct answer: C. (6) and (7)

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

Correct answer: B. ( \sqrt{14} ) and (1)

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

Correct answer: C. (3)

Explanation: 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.

Correct answer: B. \(\sqrt{12}\) चरण

Explanation: 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.

Correct answer: C. \(2\sqrt{10}\)

Explanation: 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.

Correct answer: A. \(3\sqrt{7}\)

Explanation: 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.

Correct answer: C. 8

Explanation: 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.

Correct answer: D. Because ( (\sqrt{n})^2+1^2=n+1 )

Explanation: 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.

Correct answer: C. \(3\sqrt{5}\)

Explanation: 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.

Correct answer: C. (10)

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

Correct answer: C. (5)

Explanation: 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.

Correct answer: D. At ( \sqrt{36} )

Explanation: 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 ).