Square root spiral

In a square root spiral, the hypotenuse following \(\sqrt{n}\) is \(\sqrt{n+1}\). Therefore, the hypotenuse after \(\sqrt{2024}\) is \(\sqrt{2025}\). Since \(2025=45^2\), \(\sqrt{2025}=45\). Option 44 is not correct because \(44^2=1936\), not 2025. Exam tip: Before simplifying a square root, check whether the number inside it is a perfect square.

Since \(7^2=49\) and \(8^2=64\), we have \(49<50<64\); hence \(7<\sqrt{50}<8\). Similarly, \(80\) lies between \(8^2=64\) and \(9^2=81\), so \(8<\sqrt{80}<9\). Option B may seem close, but \(\sqrt{80}>8\). Exam tip: locate a square root by comparing the number with nearby perfect squares.

A square-root spiral repeatedly uses the Pythagorean theorem. If the existing radius or hypotenuse has length sqrt{41} and a perpendicular segment of length 1 is added, the new hypotenuse has square equal to the sum of the two squared lengths. Therefore, (sqrt{41})^2+1^2=41+1=42, so the new length is sqrt{42}. The essential point is that lengths are combined through their squares, not by simply adding the lengths.

Thus option A gives the correct construction and equation. Option B is a common error because sqrt{41}+1 is not generally equal to sqrt{42}. Option C uses a perpendicular of length 2, which would give 45 under the square-root sign, and option D uses multiplication rather than the Pythagorean relation. Hence A follows directly from the theorem.

\(43^2=1849\) and \(44^2=1936\). Since \(1849<1935<1936\), taking square roots gives \(43<\sqrt{1935}<44\). Although it is very close to \(44\), \(1935<1936\), so its square root must be less than \(44\). Exam tip: To find the interval of a square root, compare the number with consecutive perfect squares.

In a square root spiral, each new right triangle has the other leg equal to \(1\). Therefore, starting from the hypotenuse \(\sqrt{288}\), the square of the next hypotenuse is \(288+1=289\). Hence, the new hypotenuse is \(\sqrt{289}=17\). The option with \(\sqrt{290}\) is incorrect because only \(1\) is added to the radicand. Exam tip: for the next hypotenuse, add \(1\) to the radicand and then check whether the result is a perfect square.

We have \(49^2=2401\) and \(50^2=2500\). Since \(2401<2499<2500\), taking square roots gives \(49<\sqrt{2499}<50\). Therefore, its position on the square root spiral is between 49 and 50. Although it is very close to 50, it cannot equal 50 because \(2499<2500\). Exam tip: locate a square root by comparing the number with consecutive perfect squares.

Since \(19^2=361\), \(20^2=400\), and \(21^2=441\), \(361<399<400\) gives \(19<\sqrt{399}<20\). Similarly, \(400<401<441\) gives \(20<\sqrt{401}<21\). Options B and C are incorrect because the numbers lie on opposite sides of 400, so their square roots lie on opposite sides of 20. Exam tip: Compare a number with the nearest perfect squares to locate its square root.

The previous hypotenuse is \(\sqrt{728}\), so \(n=728\). In a square root spiral, the next hypotenuse is \(\sqrt{n+1}\). Therefore, it is \(\sqrt{728+1}=\sqrt{729}\). \(\sqrt{728}\) is the previous hypotenuse, while \(\sqrt{727}\) represents an earlier step. Exam tip: add 1 to the number inside the square root to obtain the next hypotenuse.

In a square root spiral, each new hypotenuse represents \(\sqrt{n}\). Here the new hypotenuse is \(\sqrt{729}\), and \(729=27^2\), so its value is \(27\). Option \(26\) is incorrect because \(26^2=676\), while \(28^2=784\). Exam tip: Remember squares of nearby whole numbers to identify perfect squares quickly.

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.

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.

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

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

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.

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.

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.

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.

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.

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.

\(99^2<9999<100^2\) and \(10000=100^2\). Therefore \(\sqrt{9999}\) is slightly less than (100).