Square root spiral

Correct answer: A. The new right triangle has legs \(\sqrt n\) and 1, so its hypotenuse is \(\sqrt{n+1}\).

Explanation: In the next right triangle, the previous hypotenuse \(\sqrt n\) is one leg and the new leg is 1. By Pythagoras, \((\sqrt n)^2+1^2=n+1\), so the hypotenuse is \(\sqrt{n+1}\). Option B wrongly adds outside the square root. Exam tip: add squares of the legs first.

Correct answer: B. \(9<\sqrt{96}<10\)

Explanation: Since \(9^2=81\) and \(10^2=100\), and \(81<96<100\), we get \(9<\sqrt{96}<10\). The interval from 8 to 9 is incorrect because its corresponding squares lie from \(64\) to \(81\). Exam tip: To locate a square root, compare the number with the nearest smaller and larger perfect squares.

Correct answer: A. The previous hypotenuse is one leg, and the other perpendicular leg is 1 unit long.

Explanation: In the spiral, the previous hypotenuse becomes one leg and a perpendicular unit leg is added. If it is \(\sqrt{n}\), then the new hypotenuse is \(\sqrt{n+1}\). Exam tip: apply Pythagoras’ theorem at each step.

Correct answer: A. The lengths of new hypotenuses may become incorrect

Explanation: A square root spiral is built step by step with right triangles. Typically, one side has a known length and a perpendicular segment of 1 unit is added. The Pythagorean theorem then gives the next hypotenuse, for example \(\sqrt{1^2+1^2}=\sqrt{2}\), and later constructions continue in the same way. The accuracy of every new length depends on the accuracy of the sides used before it.

If the 1-unit perpendicular is not measured correctly, the triangle does not have the intended dimensions. Its hypotenuse will therefore not have the intended square-root length, and every later triangle based on that length may also be inaccurate. Thus option A is correct. The measurement error does not change the Pythagorean theorem or remove the need for a right angle; it only makes the construction’s lengths incorrect.

Correct answer: B. Irrational number and between (5) and (6)

Explanation: (28) is not a perfect square and (5^2<28<6^2). Therefore (\sqrt{28}) is irrational and lies between (5) and (6).

Correct answer: A. Because (1^2+1^2=2)

Explanation: In a right triangle the square of the hypotenuse equals the sum of squares of the sides. Therefore the hypotenuse is (\sqrt{2}).

Correct answer: B. Between 10 and 11

Explanation: \(10^2=100\) and \(11^2=121\). Since \(100<108<121\), we get \(10<\sqrt{108}<11\). Therefore, \(\sqrt{108}\) lies between 10 and 11 on the number line. It cannot lie between 9 and 10, because numbers in that interval have squares less than 100. Exam tip: Compare the number with consecutive perfect squares to find the interval of its square root.

Correct answer: A. \(\sqrt{100}=10\) and \(\sqrt{101}\) lies between \(10\) and \(11\)

Explanation: Since \(100=10^2\), \(\sqrt{100}=10\). Also, \(10^2=100<101<121=11^2\), so \(\sqrt{101}\) lies between \(10\) and \(11\) and is not a whole number. Option C may seem close, but \(\sqrt{101}=11\) would require \(101=121\). Exam tip: To locate a square root, compare the number with nearby perfect squares.

Correct answer: A. (\sqrt{58}), between (7) and (8)

Explanation: The spiral advances from \(\sqrt{n}\) to \(\sqrt{n+1}\) by adding a perpendicular side of length 1. Therefore, to construct \(\sqrt{59}\), the preceding hypotenuse must be \(\sqrt{58}\). The numerical position is found by comparing 59 with nearby perfect squares: \(7^2=49\) and \(8^2=64\).

Since \(49<59<64\), taking positive square roots gives \(7<\sqrt{59}<8\). Thus the correct pair is \(\sqrt{58}\) and the interval between 7 and 8, as stated in option A. \(\sqrt{57}\) would lead to \(\sqrt{58}\), while \(\sqrt{60}\) is a later length. The interval 8 to 9 is also too high.

Correct answer: B. Between 12 and 13

Explanation: We have \(12^2=144\) and \(13^2=169\). Since \(145\) is greater than \(144\) but less than \(169\), \(12<\sqrt{145}<13\). Hence, \(\sqrt{145}\) lies between 12 and 13. The interval between 11 and 12 is incorrect because numbers in that interval have squares less than 144. Exam tip: Compare the number with the nearest perfect squares to locate its square root.

Correct answer: A. Each new unit segment should be drawn perpendicular to the previous hypotenuse at its outer endpoint.

Explanation: Each new unit segment is drawn perpendicular to the previous hypotenuse at its outer endpoint. If the earlier hypotenuse is \(\sqrt n\), the new one is \(\sqrt{(\sqrt n)^2+1^2}=\sqrt{n+1}\). Drawing repeatedly from the initial point breaks the spiral. Exam tip: check the right-angle mark.

Correct answer: B. Irrational number, between 3 and 4

Explanation: The governing concepts are the classification of square roots and comparison with consecutive perfect squares. The number 12 is not a perfect square, so √12 is not a whole number. In fact, the square root of a positive integer that is not a perfect square is irrational, so √12 is irrational. To locate it on the number line, compare 12 with 3² and 4²: 3²=9 and 4²=16. Since 9<12<16, taking positive square roots gives 3<√12<4. Therefore option B is correct. It cannot be at 3 because 3²=9, and it cannot be at 4 because 4²=16. Option C places it in the wrong interval, while A and D wrongly call it a whole number.

Correct answer: C. \(\sqrt{9}\)

Explanation: For the new right triangle, hypotenuse² = \((\sqrt{8})^2+1^2=8+1=9\), so the hypotenuse is \(\sqrt{9}\). \(\sqrt{10}\) would follow \(\sqrt{9}\), not \(\sqrt{8}\). Exam tip: add 1 at each spiral step.

Correct answer: A. Adding a 1-unit perpendicular to the previous hypotenuse

Explanation: The governing concept is the geometric construction rule of the square-root spiral, not ordinary addition of lengths. At each stage, the previous hypotenuse is retained as one leg of a new right triangle, and a perpendicular segment of length 1 unit is drawn as the second leg. Pythagoras then determines the new hypotenuse. Starting with √1, we get (√1)²+1²=1+1=2, so the new hypotenuse is √2. Starting with √2, we get (√2)²+1²=2+1=3, so the next one is √3. Thus option A describes the common operation. Option D is a tempting but incorrect arithmetic shortcut: √1+1 is not √2 and √2+1 is not √3. The old hypotenuse is neither doubled nor erased.

Correct answer: B. Between 8 and 9

Explanation: \(8^2=64\) and \(9^2=81\). Since \(64<76<81\), we get \(8<\sqrt{76}<9\). Hence, the new hypotenuse in the square root spiral lies between 8 and 9. A close distractor may seem tempting because 76 is near 81, but it is still less than 9 when square-rooted. Exam tip: Compare the number with nearby perfect squares to locate its square root.

Correct answer: D. Making the hypotenuse by directly adding (1) to the previous hypotenuse

Explanation: In the square-root spiral, every new triangle is right-angled. The previous hypotenuse is used as one leg, a perpendicular segment of length 1 is added, and the new hypotenuse is calculated with Pythagoras. Thus, if the old length is \(\sqrt{n}\), the new length is \(\sqrt{n+1}\), not a simple arithmetic sum.

Directly adding 1 to the previous hypotenuse is therefore incorrect. For example, from \(\sqrt{8}\), the next length is \(\sqrt{8+1}=\sqrt{9}\), not \(\sqrt{8}+1\). Options A, B, and C describe essential parts of the construction: making a right angle, using a unit perpendicular, and reusing the previous hypotenuse. Hence option D is the most incorrect.

Correct answer: A. \(\sqrt{10}\) lies between 3 and 4.

Explanation: Since \(3^2=9\) and \(4^2=16\), we have \(9<10<16\), so \(3<\sqrt{10}<4\). The value inside the root is not compared directly with 4. Exam tip: bracket a square root using the nearest perfect squares.

Correct answer: B. The statement is correct because the new hypotenuse will be \(\sqrt{7+1}=\sqrt{8}\).

Explanation: The statement is correct. With previous hypotenuse \(\sqrt{7}\) and a new perpendicular of 1, Pythagoras gives hypotenuse² = 7 + 1 = 8, so the new hypotenuse is \(\sqrt{8}\). Exam tip: add squares, not the lengths themselves.

Correct answer: A. So that ((\sqrt{n})^2+1^2=n+1) can apply at each step

Explanation: With a (1) unit perpendicular and a (90^\circ) angle, Pythagoras theorem applies correctly. This forms successive square roots.

Correct answer: A. The hypotenuses will not be \(\sqrt{1},\sqrt{2},\sqrt{3},\ldots\) in order.

Explanation: In a standard square root spiral, each new perpendicular is 1 cm, so \(h^2\) increases by 1 at every step. With a 2 cm side, \(h^2\) increases by 4 instead. Exam tip: always check the unit perpendicular.

Correct answer: A. The previous triangle’s hypotenuse and the newly added unit-length side

Explanation: At each step, a new unit segment is drawn perpendicular to the previous hypotenuse. If that hypotenuse is \(\sqrt{n}\), the next one becomes \(\sqrt{n+1}\). In exams, identify the newly formed outer right triangle.

Correct answer: A. \(\sqrt{2}, \sqrt{3}, \sqrt{4}, \ldots\)

Explanation: Each new right triangle in the spiral is formed by adding a side of length 1. By Pythagoras’ theorem, successive hypotenuse squares are 2, 3, 4, …, so A is correct. Exam tip: check the sequence of squared lengths first.

Correct answer: B. Irrational number

Explanation: Since 17 is not a perfect square, \(\sqrt{17}\) cannot be written as a ratio of two integers. Hence, its point on the spiral represents an irrational number. Exam tip: only square roots of perfect squares are integers.

Correct answer: B. (\sqrt{254})

Explanation: Adding a (1) unit perpendicular to (\sqrt{254}) gives (\sqrt{255}). The previous hypotenuse has one less number.

Correct answer: A. A successive chain of right triangles where previous hypotenuse and (1) unit perpendicular form the next square root

Explanation: The rule of the square root spiral is based on Pythagoras theorem and a (1) unit perpendicular. This forms successive square roots.