Square root spiral

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.

Hypotenuses are formed successively in the spiral. In the usual construction, intermediate square roots are not skipped.

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 next hypotenuse is (\sqrt{m+1}). If (m+1) is a perfect square, its square root will be a whole number.

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 next hypotenuse is (\sqrt{49}), and (\sqrt{49}=7). At a perfect square, the hypotenuse becomes a whole number.

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.

The previous hypotenuse is \(\sqrt{224}\), so \(n=224\). In a square root spiral, the next hypotenuse is \(\sqrt{n+1}\); hence it is \(\sqrt{224+1}=\sqrt{225}\). \(\sqrt{224}\) is the previous hypotenuse itself, while \(\sqrt{223}\) belongs to the preceding step. Exam tip: for the next hypotenuse, add 1 to the number inside the square root.

In a square root spiral, each new hypotenuse represents \(\sqrt{n}\). Here the new hypotenuse is \(\sqrt{225}\), and since \(225=15^2\), \(\sqrt{225}=15\). Option 14 is incorrect because \(14^2=196\), while \(16^2=256\). Exam tip: first check whether the number under the square root is a perfect square.