Square root spiral

Correct answer: A. √81 = 9

Explanation: The governing concept is the Pythagorean theorem applied to the successive right triangles of a square root spiral. The old hypotenuse has length √80, and the newly drawn perpendicular has length 1. Thus, if h is the new hypotenuse, h² = (√80)² + 1² = 80 + 1 = 81. Since a length is positive, h = √81 = 9. Therefore option A is correct. Option B incorrectly subtracts 1 from the radicand. Option C doubles 80 without any geometric justification, while option D adds 2 rather than adding the square of the new perpendicular. The next point consequently represents the exact length 9.

Correct answer: A. The new hypotenuse will be \(\sqrt{18}\); to obtain \(\sqrt{19}\), the previous segment must have length \(\sqrt{18}\).

Explanation: By Pythagoras’ theorem, the square of the new hypotenuse is \((\sqrt{17})^2+1^2=17+1=18\), so it is \(\sqrt{18}\), not \(\sqrt{19}\). To construct \(\sqrt{19}\), start with \(\sqrt{18}\). Exam tip: add 1 to the radicand at each step.

Correct answer: C. We should write (\sqrt{(\sqrt{12})^2+1^2}=\sqrt{13})

Explanation: Lengths are not added directly in a square root spiral. The correct method is to add squares using Pythagoras theorem.

Correct answer: D. (\sqrt{144}), exactly at (12)

Explanation: The new hypotenuse is (\sqrt{144}). Since (\sqrt{144}=12), it is not in an interval but exactly at (12).

Correct answer: A. (\sqrt{34}) and (1)

Explanation: In a square root spiral, each new right triangle has a perpendicular side of length 1. The new hypotenuse is obtained from the previous hypotenuse by applying the Pythagorean theorem. Thus, to obtain the length \\(\\sqrt{35}\\), we must begin with a previous hypotenuse whose square is one less than 35. This makes option A the appropriate choice.

Using the theorem, the required calculation is \\( (\\sqrt{34})^2+1^2=34+1=35 \\). Therefore, the new hypotenuse is \\(\\sqrt{35}\\). A previous length of \\(\\sqrt{33}\\) with a side of 2 would also give 37, not 35, so option B does not fit this construction. The other choices use the target or an excessive previous value.

Correct answer: A. पिछले कर्ण के एक सिरे पर 1 इकाई का लंब खींचकर बने नए समकोण त्रिभुज के कर्ण के रूप में

Explanation: Each new right triangle has the previous hypotenuse \(\sqrt{n-1}\) as one side and a perpendicular side of length 1. By Pythagoras, its hypotenuse is \(\sqrt{(n-1)+1}=\sqrt n\). Exam tip: track successive hypotenuses.

Correct answer: A. At each new stage, a side of length 1 unit is drawn perpendicular to the previous hypotenuse at its endpoint.

Explanation: The previous hypotenuse \(\sqrt{n}\) and the new unit side form a right angle. By Pythagoras, the next hypotenuse is \(\sqrt{n+1}\), not \(\sqrt{n^2+1}\). Exam tip: first locate the right-angle mark.

Correct answer: B. (\sqrt{225}), between (15) and (16)

Explanation: The direct answer is B. In the usual square-root spiral, each new squared hypotenuse increases by 1. Therefore, immediately before sqrt{226} comes sqrt{225}. To locate the value, compare 226 with neighbouring perfect squares: 15^2=225 and 16^2=256, so 15^2<226<16^2 and hence 15<sqrt{226}<16. Option A is wrong because sqrt{224} is not the immediately preceding term and its interval is also not the stated one. Option B correctly gives both the previous hypotenuse and the interval. Option C repeats the new hypotenuse instead of the previous one. Option D gives a later value and an incorrect interval. Exam cue: compare the radicand with consecutive squares.

Correct answer: A. √169 is formed, and 169 is a perfect square

Explanation: The relevant rule is that adding a perpendicular of length 1 at a right angle increases the square of the current hypotenuse by 1. Starting with a hypotenuse √168, the next one is therefore √((√168)² + 1²) = √(168 + 1) = √169. Since 169 = 13 × 13 = 13², the radical simplifies exactly to the whole number 13. Hence option A is correct. Option B results from subtracting 1, which is not the rule of the construction. Option C doubles the radicand without using the Pythagorean theorem, and option D wrongly assumes that the added perpendicular causes no change. The special simplification occurs because 169 is a perfect square.

Correct answer: B. \(\sqrt{242}\) is between 15 and 16, and \(\sqrt{256}=16\)

Explanation: Since \(15^2=225\) and \(16^2=256\), and \(225<242<256\), we get \(15<\sqrt{242}<16\). Also, \(\sqrt{256}=16\) because \(256=16^2\). Therefore, option B is correct. Option D is incorrect because \(\sqrt{242}\) is less than 16. Exam tip: To locate a square root, compare the number with the nearest perfect squares on either side.

Correct answer: A. \(\sqrt{4},\ \sqrt{9},\ \sqrt{16}\)

Explanation: In a square root spiral, a point’s distance from the origin is its hypotenuse label \(\sqrt{n}\). This is an integer only when \(n\) is a perfect square; for example, \(\sqrt{9}=3\). Option B fails because \(\sqrt{12}\) is not an integer. Exam tip: look for perfect-square radicands.

Correct answer: B. \(11^2<125<12^2\)

Explanation: \(11^2=121\) and \(12^2=144\). Since \(121<125<144\), the correct inequality is \(11^2<125<12^2\), and \(\sqrt{125}\) lies between 11 and 12. \(10^2<125<11^2\) is incorrect because \(11^2=121\), which is less than 125. Exam tip: To locate a square root, compare the number with the nearest perfect squares.

Correct answer: A. It is 7 units from the origin.

Explanation: Since \(49=7^2\), \(\sqrt{49}=7\), so its point is 7 units from the origin. It is not irrational. Exam tip: square roots of perfect squares are integers.

Correct answer: A. ((\sqrt{7})^2+1^2=8)

Explanation: By Pythagoras theorem, the squares of sides are added. Therefore (\sqrt{7}) and (1) form hypotenuse (\sqrt{8}).

Correct answer: A. The next hypotenuse is (\sqrt{64}=8), and (\sqrt{65}) is between (8) and (9)

Explanation: After (\sqrt{63}), (\sqrt{64}=8) is formed. Since (8^2<65<9^2), (\sqrt{65}) lies between (8) and (9).

Correct answer: B. Always a whole number

Explanation: The governing construction principle is that a unit perpendicular is added at a right angle, changing a hypotenuse √m into √(m+1). The condition says that m is exactly one less than a perfect square. Thus there is some whole number r such that m+1 = r². The next hypotenuse is therefore √(m+1) = √(r²) = r, which is a whole number. Option B is correct. It is not always irrational, as option A claims, because the condition deliberately makes the new radicand a perfect square. Option C has no relation to the triangle, and option D ignores the increase caused by the new perpendicular.

Correct answer: B. (16<\sqrt{288}<17)

Explanation: Because (16^2=256) and (17^2=289). The number (288) lies between them, so (\sqrt{288}) lies between (16) and (17).

Correct answer: A. Because in the usual rule the new perpendicular is (1) unit and the previous hypotenuse should be (\sqrt{47})

Explanation: In the usual square root spiral, a (1) unit perpendicular is added every time. For (\sqrt{48}), the previous hypotenuse is (\sqrt{47}).

Correct answer: A. \(\sqrt{26}\) lies between 5 and 6; in the spiral, it is represented by the hypotenuse labelled \(\sqrt{26}\).

Explanation: Since \(25<26<36\), we get \(5<\sqrt{26}<6\). A nearby perfect square does not make the root equal to 5. In the spiral, each new hypotenuse represents its corresponding root. Exam tip: bracket roots using consecutive perfect squares.

Correct answer: A. \(\sqrt{8}\rightarrow\sqrt{9}\rightarrow\sqrt{10}\)

Explanation: The hypotenuses increase in order in the spiral. Just before \(\sqrt{10}\) comes \(\sqrt{9}\), and before that \(\sqrt{8}\).

Correct answer: B. Mistakenly placing (\sqrt{195}) between (14) and (15)

Explanation: Actually (13^2=169) and (14^2=196). Therefore (\sqrt{195}) lies between (13) and (14), less than (14).

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

Explanation: In Pythagoras theorem, the square of the hypotenuse equals the sum of squares of the sides. This is the general rule.

Correct answer: A. \(\sqrt{196}\)

Explanation: Since \(196=14^2\), \(\sqrt{196}=14\), which is rational. The numbers 198, 200 and 202 are not perfect squares, so their square roots are irrational. Exam tip: check for a perfect square first.

Correct answer: A. A perfect square number

Explanation: \(\sqrt{n}\) is rational only when \(n\) is a perfect square. For example, \(\sqrt{49}=7\), so that spiral segment has rational length. Being prime or odd is not sufficient. Exam tip: check for perfect squares first.

Correct answer: A. पिछले कर्ण के वर्ग में 1 जोड़ने के वर्गमूल के बराबर

Explanation: Let the previous hypotenuse be \(\sqrt{n}\). With a new perpendicular side of 1, Pythagoras gives \(\sqrt{(\sqrt{n})^2+1^2}=\sqrt{n+1}\). It is not obtained by simply adding 1 to the length. Exam tip: add the squares of perpendicular sides.