Square root spiral

In a square root spiral, the hypotenuse is formed by the sum of squares. Do not treat (\sqrt{18}+1) as (\sqrt{19}).

The existing hypotenuse is \(\sqrt{n}\), and the new perpendicular side is \(3\) units. By the Pythagorean theorem, the square of the new hypotenuse is \((\sqrt{n})^2+3^2=n+9\). Therefore, the new hypotenuse is \(\sqrt{n+9}\). \(\sqrt{n+3}\) is incorrect because the perpendicular contributes \(3^2=9\), not \(3\). Exam tip: while finding a hypotenuse, add the squares of the two perpendicular sides.

In a square root spiral, the next 1-unit side is drawn perpendicular to the previous hypotenuse, making a right triangle. Hence the new hypotenuse satisfies \(h^2=a^2+1\). Exam tip: always check the perpendicular condition.

The \(\sqrt{n}\) segment and a perpendicular unit segment are the legs. Pythagoras gives hypotenuse squared \(=n+1\), hence \(\sqrt{n+1}\). Exam tip: verify the right angle.

The square root spiral follows the rule that adding a 1-unit perpendicular at a right angle increases the squared hypotenuse by 1. Starting from √143, the next hypotenuse is therefore √(143 + 1) = √144. Since 144 = 12², its principal square root is exactly 12. Thus option B is correct. Option A corresponds to √121 and is too small; option C corresponds to √169 and is too large; and option D is false because the next radicand is a perfect square and therefore produces a whole-number length. The calculation is exact and does not require a decimal approximation.

The stated rule says that the kth hypotenuse has length \\(\\sqrt{k}\\). To identify the hypotenuse represented by \\(\\sqrt{36}\\), read 36 as the index and then evaluate the radical to find its length. Thus it is the 36th hypotenuse, and its numerical length is 6. Option B is therefore correct.

Since \\(36=6^2\\), we have \\(\\sqrt{36}=6\\). The number 36 must be retained as the position, while 6 is the measured value of the hypotenuse. Option D incorrectly gives 36 as the length. The 35th and 37th hypotenuses would instead have lengths \\(\\sqrt{35}\\) and \\(\\sqrt{37}\\), so neither option A nor C matches the required radical.

Only the newly added perpendicular has length 1; the other leg is the previous hypotenuse. For example, after a hypotenuse of \(\sqrt{2}\), the next one is \(\sqrt{3}\), so the triangles are not congruent. Exam tip: track the changing hypotenuse.

Each new right triangle has one leg of length 1 and the previous hypotenuse as the other leg. By Pythagoras, if the previous square is n, the next is n+1. Option B wrongly adds 1 to lengths. Exam tip: track squared lengths first.

\(16^2=256\) and \(17^2=289\). Since \(256<257<289\), the correct inequality is \(16^2<257<17^2\), so \(\sqrt{257}\) lies between 16 and 17. The close distractor \(15^2<257<16^2\) is incorrect because \(16^2=256\), which is less than 257. Exam tip: compare the number with nearby perfect squares to locate its square root.

In a square root spiral, the hypotenuse after \(\sqrt{n}\) is \(\sqrt{n+1}\). Since the given hypotenuse is 25, \(\sqrt{n+1}=25\). Squaring both sides gives \(n+1=625\), so \(n=624\). Option 625 is incorrect because it is the value of \(n+1\), not \(n\). Exam tip: square the given hypotenuse and then check the one-step change in the spiral sequence.

The governing concept is the successive construction of square roots by right triangles. Starting with a unit segment and a perpendicular unit segment gives a hypotenuse of √2. In the next step, that hypotenuse is used as one leg and another unit perpendicular segment is drawn, producing √3. Repeating the same process produces √4 and then √5. Therefore the order is √2, √3, √4, √5, so option B is correct. Option A skips √3, while C and D do not follow the forward, successive construction. The spiral is based on the Pythagorean theorem, not on an arbitrary ordering of roots.

Since \(13^2=169\) and \(14^2=196\), we have \(169<170<196\) and \(169<195<196\). Therefore, both \(\sqrt{170}\) and \(\sqrt{195}\) lie between \(13\) and \(14\). In particular, \(\sqrt{195}\) is slightly less than \(14\) because \(195<196\), so it cannot lie between \(14\) and \(15\). Exam tip: locate a square root by comparing its radicand with nearby perfect squares.

After (\sqrt{24}), (\sqrt{25}=5) is formed. Since (5^2<26<6^2), (\sqrt{26}) lies between (5) and (6).

The governing rule of the spiral is that the next hypotenuse after √m is √(m+1), because a new perpendicular unit segment is added and the Pythagorean theorem gives (new hypotenuse)²=m+1. If m is exactly one less than a perfect square, write m+1=k² for some whole number k. Then √(m+1)=√(k²)=k, which is a whole number. Thus option A is correct. The value is not necessarily irrational, so B is too strong; it is not zero, so C is false; and the new step changes the radicand, so it does not remain √m as claimed in D.

Since \(20^2=400\) and \(21^2=441\), we have \(400<440<441\). Therefore, \(20<\sqrt{440}<21\), whereas \(\sqrt{441}=\sqrt{21^2}=21\). Option B is incorrect because \(440<441\), so \(\sqrt{440}\) cannot be 21. Exam tip: compare a number with the nearest perfect squares to locate its square root.

The governing construction rule says that after the hypotenuse √n, the next hypotenuse is √(n+1). Therefore, after √48, the next hypotenuse is √49. Since 49 is a perfect square, √49=7, so option B is correct. Option A moves in the wrong direction and represents √47. Option C incorrectly doubles the radicand, while option D adds two rather than one to 48. The important calculation is not an approximation: √48 is followed by √49 exactly, and the perfect-square result makes the next length the whole number 7. This illustrates how the spiral can represent both irrational and integral square roots.

We have \(24^2=576\) and \(25^2=625\). Since \(576<624<625\), taking positive square roots gives \(24<\sqrt{624}<25\). It cannot lie between 25 and 26 because \(624<25^2\). Exam tip: compare the number with consecutive perfect squares to locate its square root.

In a square root spiral, the hypotenuse after \(\sqrt{899}\) is \(\sqrt{900}\). Since \(900=30^2\), \(\sqrt{900}=30\). Although 29 is close, \(29^2=841\), not 900. Exam tip: when the number under a square root is a perfect square, its square root is a whole number.

Since \(5^2=25\) and \(6^2=36\), and \(25<27<36\) as well as \(25<32<36\), we get \(5<\sqrt{27}<6\) and \(5<\sqrt{32}<6\). Hence, both lengths lie between 5 and 6 in the square root spiral. Option B is wrong because \(\sqrt{27}\) is not between 4 and 5. Exam tip: To locate a square root, compare the number with the nearest perfect squares.

The square root spiral uses the Pythagorean theorem. When a right triangle has one leg equal to the previous hypotenuse \(\sqrt{7}\) and the newly added perpendicular equal to 1, the square of the new hypotenuse is the sum of the squares of these two legs. Therefore, \(h^2=(\sqrt{7})^2+1^2=7+1=8\), and the new hypotenuse is \(h=\sqrt{8}\), taking the positive length.

Hence option A gives the correct reason. The expression \(\sqrt{7}+1\) is not generally equal to \(\sqrt{8}\), because lengths are not added directly in this construction. Multiplication is also irrelevant, and using 2 instead of 1 would give \(7+4=11\), not 8. The supplied answer correctly applies the theorem and is fully consistent with the spiral construction.

\(30^2=900\) and \(31^2=961\). Since \(900<960<961\), taking square roots gives \(30<\sqrt{960}<31\). Therefore, option B is correct. Although 960 is very close to 961, it is still less than 961, so \(\sqrt{960}\) is less than 31. Exam tip: To locate a square root, find the consecutive perfect squares on either side of the given number.

The rule \(\sqrt{n}\) to \(\sqrt{n+1}\) works only when the new segment has length 1 and is drawn perpendicular to the existing hypotenuse. The perpendicular condition creates a right triangle, allowing Pythagoras to be used. Without a right angle, the simple sum of squares does not follow from the given construction.

If the old hypotenuse is \(\sqrt{n}\), then \(h^2=(\sqrt{n})^2+1^2=n+1\), so \(h=\sqrt{n+1}\). Therefore option A contains both necessary conditions. A segment of length 2 would produce \(\sqrt{n+4}\), direct addition is not valid, and an equilateral triangle is not the required shape.

In a square root spiral, each new hypotenuse is obtained by adding 1 to the number under the previous radical. Thus, after \(\sqrt{168}\), the next hypotenuse is \(\sqrt{168+1}=\sqrt{169}\). Since \(169=13^2\), \(\sqrt{169}=13\). Option B has the correct radicand but gives the wrong value. Exam tip: add 1 to the radicand for the next hypotenuse, then check whether it is a perfect square.

\(28^2=784\) and \(29^2=841\). Since \(784<840<841\), taking square roots gives \(28<\sqrt{840}<29\). Therefore, on the square root spiral, \(\sqrt{840}\) lies between \(\sqrt{784}=28\) and \(\sqrt{841}=29\). The interval after \(29\) is incorrect because \(840<841\). Exam tip: compare a number with consecutive perfect squares to locate its square root quickly.

Since \(7^2=49\) and \(8^2=64\), we have \(49<50<64\) and \(49<63<64\). Hence, \(7<\sqrt{50}<8\) and \(7<\sqrt{63}<8\). Therefore, both points lie between 7 and 8 on the square root spiral. Option A is incorrect because \(50>49=7^2\), so \(\sqrt{50}\) cannot lie between 6 and 7. Exam tip: To locate a square root, compare the number with consecutive perfect squares.