Square root spiral

The governing concept is the Pythagorean theorem. The two perpendicular sides are √3 and 1, so their squared lengths give (√3)² + 1² = 3 + 1 = 4, and the hypotenuse is √4. However, √3 + 1 is a direct sum of side lengths and is not equal to √4; numerically it is about 2.732, whereas √4 is 2. Hence option C contains the logical error.

Use the square-root spiral rule and convert the given whole-number hypotenuse into radical form. Since √(n+1) = 12, squaring both sides gives n + 1 = 144, so n = 143. The previous hypotenuse was therefore √n = √143. Option B is correct; √144 is the new hypotenuse itself, while √142 and √145 do not satisfy the equation.

Answer: option C, √729 = 27. In the spiral, the next hypotenuse is obtained by adding a perpendicular unit segment. If the present hypotenuse is √728, then the next length h satisfies h² = (√728)² + 1² = 728 + 1 = 729. Since 729 = 27², h = √729 = 27 exactly. Therefore the point transferred to the number line is exactly 27, not merely a value close to it. Options A and B incorrectly place √729 below 27, while option D places it above 27. The key step is checking whether the radicand is a perfect square. Memory cue: 27² = 729, so √729 simplifies exactly to 27.

The governing idea is the indexing rule of the square-root spiral: the kth hypotenuse has length √k. Therefore, an expression √144 corresponds directly to k = 144, so it is the 144th hypotenuse. Its numerical length is found by identifying the positive number whose square is 144. Since 12 × 12 = 144, √144 = 12. Thus both parts of the answer must be correct: the position is 144th and the length is 12. Option A has the wrong index, while option B also shifts the index incorrectly. Option C uses 144 as the value of the square root, confusing the radicand with its square root. Hence option D is the only complete and correct answer.

In the square-root spiral, adding the next unit perpendicular changes a hypotenuse √n into √(n + 1), by the Pythagorean theorem. Starting with √255, the next hypotenuse is therefore √(255 + 1) = √256. Since 256 is a perfect square and 16 × 16 = 256, √256 = 16. Thus the exact value of the next hypotenuse is 16. Option A moves backward to √254, not forward. Option C incorrectly doubles 255 rather than adding one, and option D adds two instead of one. The perfect-square result is why the radical simplifies completely. Hence option B is the only correct answer.

The governing concept is the Pythagorean theorem used in the successive right triangles of a square-root spiral. Since PQ is perpendicular to OP, triangle OPQ is right-angled at P, and OQ is the hypotenuse. Therefore, OQ² = OP² + PQ² = (√35)² + 1² = 35 + 1 = 36. Taking the positive square root, because a geometric length cannot be negative, gives OQ = √36 = 6. Hence option C is correct because it gives the exact radical form requested. Option B merely repeats the old hypotenuse, option A incorrectly subtracts the squares, and option D gives OQ² rather than OQ itself.

The governing pattern is the recursive construction of the square-root spiral. The first right triangle has two unit perpendicular sides, so its hypotenuse is √2. At every later stage, a perpendicular segment of length 1 is added to the current hypotenuse. Thus, if the rth triangle is counted from the beginning, its hypotenuse H satisfies H² = r + 1, or H = √(r + 1). For r = 18, H₁₈ = √(18 + 1) = √19. Therefore option C is correct. Option A is one stage too small, option B incorrectly treats the first triangle as √1, and option D is one stage too large. The answer is an exact radical, so no decimal approximation is needed.

The governing concept is classification of numbers after simplifying a square root. Factor 50 as 25 × 2. Therefore √50 = √(25×2) = 5√2. The number √2 is irrational because it cannot be expressed as a ratio of two integers; multiplying it by the non-zero integer 5 remains irrational. Hence OP = 5√2 is irrational, so option C is correct. It is not rational, and consequently it cannot be an integer or a natural number. A common error is to see 50 as an integer under the radical and assume its root must also be an integer. Only perfect-square radicands produce integer square roots; 50 is not a perfect square.

The governing idea is the recursive rule of the square-root spiral: every new right triangle contributes a perpendicular unit segment, so the square of the hypotenuse increases by 1 at each step. The current hypotenuse is OP = √8, which means OP² = 8. Five new steps therefore change the squared length by 5, giving 8 + 5 = 13. The final hypotenuse is consequently √13. Option C is correct. √11 would correspond to only three new steps, √12 to four steps, and √14 to six steps. The calculation concerns the square of the length first; only after adding the contributions do we take the square root to obtain the final length.

The governing concept is the Pythagorean theorem in one step of a square-root spiral. If the existing hypotenuse is √26 and the newly constructed perpendicular segment has length 1, the square of the next hypotenuse equals the sum of the squares of these perpendicular parts. Hence (√26)² + 1² = 26 + 1 = 27, so the new hypotenuse is √27. Option A is therefore correct. In option B, the added segment is incorrectly taken as 2, giving 26 + 4 = 30. Option C starts with √27 and produces 28 after adding 1, while option D starts with √25 and produces 26. Thus only A represents the correct spiral step.

In a square root spiral, each newly drawn perpendicular segment increases the number under the square root by 1: √40, √41, √42, and so on. The number of steps from 40 to 47 is 47 − 40 = 7. Therefore seven new perpendicular segments are required, making option C correct. Counting the endpoints or subtracting the wrong way would produce the distractor values.

The governing idea is the indexing rule of the square-root spiral: the point with index n is at distance OPₙ = √n from the origin. Therefore, OPₖ₊₁ must also equal √(k + 1). The question gives OPₖ₊₁ = √17, so we equate the radicands: √(k + 1) = √17. Since both sides are non-negative, their radicands are equal, giving k + 1 = 17 and hence k = 16. Thus option D is correct. Option A incorrectly uses 17 as k itself, while options B and C result from subtracting or adding the wrong amount. The important point is that the subscript k + 1 represents the next index, not the value of the distance.

The governing concept is the Pythagorean theorem, because the new side is drawn perpendicular to the existing hypotenuse-like segment. The two perpendicular legs are 5 units and 1 unit, so if L is the new hypotenuse, then L² = 5² + 1² = 25 + 1 = 26. Taking the positive square root, as lengths are positive, gives L = √26 units. Therefore option A is correct. Option B, 6, comes from adding the lengths directly, but perpendicular sides combine through squares, not ordinary addition. Option C uses an incorrect subtraction, and option D is only 5², omitting the contribution of the new unit side. This repeated Pythagorean step is what generates successive square-root distances in the spiral.

The defining construction rule is repeated at every stage: a new segment of length 1 unit is drawn perpendicular to the previous radius or hypotenuse, and the resulting hypotenuse gives the next square-root distance. Thus, when moving through the constructions for √3 and √4, the newly added outer side is 1 unit and perpendicular in each case. Option B is therefore correct. The old radius is not 1 in both steps; it changes from √2 to √3. The hypotenuses are √3 and √4 = 2, so they are not equal. The triangle areas also differ because their corresponding previous sides differ. The common feature concerns the construction step, not the numerical values of the old radius, hypotenuse, or area.

The governing concept is the Pythagorean update used in the square-root spiral. If the current hypotenuse is √6 and a new side of length 1 is drawn perpendicular to it, the next hypotenuse L satisfies L² = (√6)² + 1² = 6 + 1 = 7. Hence L = √7, not √6 + 1. The student has added lengths directly, but perpendicular components must be combined by adding their squares. Option A is correct. Option B confuses the squared length with the length itself; 7 is L², not L. Option C ignores the new segment, and option D decreases the distance without mathematical justification. This distinction is central to the spiral’s construction and prevents the common error of treating a right-triangle step like ordinary linear addition.

The construction begins with the existing distance √2, which acts as one leg of a right triangle. A new segment of length 1 unit is drawn perpendicular to it, so the two legs are √2 and 1. By the Pythagorean theorem, the new hypotenuse has length √[(√2)² + 1²] = √(2 + 1) = √3. Therefore option A correctly identifies the triangle. Option B replaces √2 by 2 and also incorrectly states that the hypotenuse is 3. Option C reverses the roles of the known distance and the new hypotenuse, while option D would produce a hypotenuse √2, not √3, because 1² + 1² = 2. The right angle and the unit step are essential features of the spiral.

The governing concept is comparison of a square root with nearby perfect squares. Since 36 = 6² and 49 = 7², and 36 < 48 < 49, taking principal square roots preserves the order and gives 6 < √48 < 7. Thus the final spiral distance √48 lies strictly between 6 and 7, so option A is correct. Option B is irrelevant: being even does not determine whether a square root lies between these particular integers. Option C gives only an upper comparison with 50 and does not establish the lower bound 6. Option D incorrectly treats a number and its square root as equal. Comparing with consecutive perfect squares is a dependable method for locating any positive square root without needing a decimal approximation.

The governing theorem is the Pythagorean theorem, which connects the three sides of a right triangle. In this construction, one leg has length √7 and the newly drawn perpendicular leg has length 1. Therefore the square of the hypotenuse is (√7)² + 1² = 7 + 1 = 8, so the hypotenuse is √8. This is precisely how successive square-root lengths are generated in the spiral. Option C is correct. Thales’ theorem concerns angles in a semicircle, the midpoint theorem concerns a segment joining midpoints, and an area theorem does not directly provide this side-length relation.