Square root spiral

Correct answer: B. \(\sqrt{1600}\)

Explanation: In a square root spiral, when one side is \(\sqrt{n}\) and the perpendicular side is 1 unit, the square of the new hypotenuse is \(n+1\). Therefore, drawing a 1-unit perpendicular on \(\sqrt{1599}\) gives \(\sqrt{1599+1}=\sqrt{1600}\). \(\sqrt{1598}\) would result from subtracting 1, while \(\sqrt{1601}\) would require an increase of 2. Exam tip: at each new step, add 1 to the number inside the square root.

Correct answer: B. \(36<\sqrt{1368}<37\)

Explanation: We have \(36^2=1296\) and \(37^2=1369\). Since \(1296<1368<1369\), taking square roots gives \(36<\sqrt{1368}<37\). It cannot equal 37 because \(37^2=1369\), not 1368. Exam tip: To locate a square root, compare the number with the nearest perfect squares on either side.

Correct answer: A. Take the (\sqrt{3}) hypotenuse length in a compass and draw an arc from the origin

Explanation: The hypotenuse length of the square root to be marked is taken in the compass. An arc from the origin gives the correct location.

Correct answer: B. (\sqrt{839})

Explanation: Drawing a (1) unit perpendicular on (\sqrt{839}) forms (\sqrt{840}). The previous hypotenuse has one less number.

Correct answer: A. \(\sqrt{960}\) lies between 30 and 31, and \(\sqrt{1024}=32\).

Explanation: Since \(30^2=900\) and \(31^2=961\), and \(900<960<961\), we get \(30<\sqrt{960}<31\). Also, \(1024=32^2\), so \(\sqrt{1024}=32\). Option B is incorrect because 960 is less than \(31^2\), so its square root cannot be greater than 31. Exam tip: Compare a number with nearby perfect squares to estimate its square root.

Correct answer: A. Because ((\sqrt{9})^2+1^2=10)

Explanation: In a square root spiral, the existing hypotenuse is used as one leg of a new right triangle, and a perpendicular segment of length 1 is added. If the existing hypotenuse is \(\sqrt{9}\), then the square of the next hypotenuse is \((\sqrt{9})^2+1^2=9+1=10\). Thus the new hypotenuse is \(\sqrt{10}\). The construction depends on adding squares, not adding the lengths themselves.

Therefore option A is correct. Option B incorrectly claims \(\sqrt{9}+1=\sqrt{10}\); in fact, \(3+1=4\), not \(\sqrt{10}\). Option C confuses the number 10 with its square root, while option D subtracts the squares and does not represent the right-triangle construction. The previous hypotenuse and the unit perpendicular work together through the Pythagorean theorem, so the supplied answer is accurate.

Correct answer: B. \(\sqrt{1849}=43\)

Explanation: In a square root spiral, successive hypotenuses are \(\sqrt{1},\sqrt{2},\sqrt{3}\), and so on. Therefore, the hypotenuse after \(\sqrt{1848}\) is \(\sqrt{1849}\). Since \(1849=43^2\), we get \(\sqrt{1849}=43\), so it lies at an integer value. \(\sqrt{1850}\) comes next, but 1850 is not a perfect square. Exam tip: add 1 to the radicand for the next hypotenuse, then check whether it is a perfect square.

Correct answer: A. \(\sqrt{35}\) lies between 5 and 6, while \(\sqrt{37}\) lies between 6 and 7

Explanation: Since \(5^2=25<35<36=6^2\), we get \(5<\sqrt{35}<6\). Similarly, \(6^2=36<37<49=7^2\), so \(6<\sqrt{37}<7\). Therefore, option A is correct. Option B places \(\sqrt{37}\) incorrectly, while option C places \(\sqrt{35}\) incorrectly. Exam tip: compare a number with consecutive perfect squares to locate its square root.

Correct answer: A. \(\sqrt{2209}=47\)

Explanation: In a square root spiral, successive hypotenuses are \(\sqrt{1}, \sqrt{2}, \sqrt{3}\), and so on. Therefore, the hypotenuse after \(\sqrt{2208}\) is \(\sqrt{2209}\). Since \(47^2=2209\), \(\sqrt{2209}=47\). Option B has the correct radicand but an incorrect value, while option C skips to the next term. Exam tip: identify nearby perfect squares to check whether a square root is an integer.

Correct answer: A. \(\sqrt{48}\) lies between 6 and 7, while \(\sqrt{50}\) lies between 7 and 8

Explanation: Since \(6^2=36\), \(7^2=49\), and \(8^2=64\), \(36<48<49\) gives \(6<\sqrt{48}<7\). Similarly, \(49<50<64\) gives \(7<\sqrt{50}<8\). Therefore, option A is correct. A common confusion is that both numbers are close to 49, but 48 is below 49 whereas 50 is above it. Exam tip: locate a square root by comparing the number with the nearest perfect squares.

Correct answer: A. √9999 lies between 99 and 100, and √10000 = 100

Explanation: The correct answer is A. Compare the radicand with consecutive squares: 99² = 9801 and 100² = 10000. Since 9801 < 9999 < 10000, taking positive square roots gives 99 < √9999 < 100. Also, 10000 = 100², so √10000 = 100 exactly. The two values are therefore not equal: √9999 is slightly smaller than 100, while √10000 reaches the integer 100. Option B ignores the strict inequality, option C reverses the perfect-square facts and misclassifies √10000, and option D contradicts the upper bound obtained from 100².

Correct answer: A. \((\sqrt{72})^2+1^2=73\), so the new hypotenuse is \(\sqrt{73}\)

Explanation: In a square root spiral, each new right triangle has the previous hypotenuse as one side and a unit length as the other side. By Pythagoras’ theorem, the next hypotenuse is \(\sqrt{(\sqrt{72})^2+1^2}=\sqrt{73}\). Option B incorrectly adds the hypotenuses directly; in general, the sum of square roots is not the square root of their sum. Exam tip: for the next hypotenuse, add \(1\) to the square of the previous hypotenuse.

Correct answer: A. \(\sqrt{2601}=51\)

Explanation: In a square root spiral, successive hypotenuses are \(\sqrt{1}, \sqrt{2}, \sqrt{3}\), and so on. Therefore, the hypotenuse after \(\sqrt{2600}\) is \(\sqrt{2601}\). Since \(2601=51^2\), we get \(\sqrt{2601}=51\). \(\sqrt{2602}\) is the hypotenuse after the next one, not the immediate next hypotenuse. Exam tip: For a number near a perfect square, first check whether it equals the square of an integer.

Correct answer: A. The next hypotenuse is (\sqrt{16}=4), and (\sqrt{17}) lies between (4) and (5)

Explanation: After (\sqrt{15}), (\sqrt{16}=4) is formed. Since (4^2<17<5^2), (\sqrt{17}) lies between (4) and (5).

Correct answer: B. (58<\sqrt{3480}<59)

Explanation: (58^2=3364) and (59^2=3481). The number (3480) lies between them, so (\sqrt{3480}) lies between (58) and (59).

Correct answer: A. \(\sqrt{3025}=55\)

Explanation: In a square-root spiral, adding a perpendicular segment of length 1 creates a right triangle. If the old hypotenuse is sqrt{3024}, the new hypotenuse is found by adding the square of the new unit side: sqrt{(sqrt{3024})^2+1^2}=sqrt{3024+1}=sqrt{3025}. Since 3025=55^2, its principal square root is 55. The positive value is used because a length cannot be negative.

Therefore option A is correct. Option B subtracts 1 instead of adding the perpendicular side's square. Option C doubles 3024, which does not follow from the theorem, and option D adds 2 rather than 1 while also claiming the incorrect equality sqrt{3026}=55. The exact radical form is sqrt{3025}, and its simplified exact value is 55.

Correct answer: A. \(\sqrt{2207}\) lies between 46 and 47, and \(\sqrt{2209}=47\)

Explanation: Since \(46^2=2116\), \(47^2=2209\), and \(2116<2207<2209\), we get \(46<\sqrt{2207}<47\). Also, \(2209=47^2\), so \(\sqrt{2209}=47\). Option D may seem close, but \(2207\) is less than \(2209\), so its square root cannot be 47. Exam tip: locate a square root by comparing the number with nearby perfect squares.

Correct answer: C. √(n+64)

Explanation: The relevant concept is the Pythagorean theorem for a right triangle. One perpendicular side has length √n and the other has length 8. If H is the new hypotenuse, then H² = (√n)² + 8² = n + 64. Since a geometric length is positive, H = √(n+64). Thus option C is correct. Option A adds 8 instead of its square, so it ignores the theorem. Option B would correspond to adding 4² rather than 8². Option D multiplies n by 8 and does not represent the sum of the squares of the perpendicular sides. The essential step is recognizing that the added side contributes 8²=64 to the radicand.

Correct answer: B. (\sqrt{4095})

Explanation: If the new hypotenuse is (\sqrt{n+1}), the previous hypotenuse is (\sqrt{n}). Therefore before (\sqrt{4096}), it was (\sqrt{4095}).

Correct answer: A. \(\sqrt{4096}=64\), next hypotenuse \(\sqrt{4097}\)

Explanation: Since \(4096=64^2\), \(\sqrt{4096}=64\) is an integer, not an irrational number. In a square root spiral, each new step adds \(1\) to the square of the previous hypotenuse. Therefore, the hypotenuse after \(\sqrt{4096}\) is \(\sqrt{4096+1}=\sqrt{4097}\), not \(\sqrt{8192}\). Exam tip: first check whether the radicand is a perfect square, then add \(1\) to the radicand for the next hypotenuse.

Correct answer: B. \(\sqrt{3599}\), (60)

Explanation: Before \(\sqrt{3600}\), \(\sqrt{3599}\) comes, and \(3600=60^2\). Therefore \(\sqrt{3600}=60\).

Correct answer: A. The conclusion is correct, because \(64<70<81\).

Explanation: Neelam’s conclusion is correct. We have \(8^2=64\) and \(9^2=81\). Since \(64<70<81\), taking square roots gives \(8<\sqrt{70}<9\). Option B is incorrect because 70 does not lie between \(7^2=49\) and \(8^2=64\). Exam tip: locate a square root by comparing the number with consecutive perfect squares.

Correct answer: A. \(\sqrt{1599}\)

Explanation: In a square root spiral, the successive hypotenuses are \(\sqrt{2}, \sqrt{3}, \sqrt{4}, \ldots\). Since the new hypotenuse is \(40=\sqrt{1600}\), the hypotenuse immediately before it is \(\sqrt{1599}\). \(\sqrt{1601}\) would be the next hypotenuse, while \(\sqrt{1598}\) is one step earlier. Exam tip: To place a whole number in this sequence, write it as \(\sqrt{n^2}\).

Correct answer: D. \(44<\sqrt{2023}<45\)

Explanation: We have \(44^2=1936\) and \(45^2=2025\). Since \(1936<2023<2025\), taking positive square roots gives \(44<\sqrt{2023}<45\). The interval between 43 and 44 is not correct because \(44^2\) is still less than 2023. Exam tip: To locate a square root, compare the number with consecutive perfect squares.

Correct answer: C. (\sqrt{(\sqrt{98})^2+1^2}=\sqrt{99})

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