Square root spiral

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.

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.

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

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

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.

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.

By Pythagoras, ((\sqrt{n})^2+8^2=n+64). So changing the perpendicular changes the usual sequence.

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

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.

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

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.

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 \(\sqrt{n^2}\).

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.

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

By Pythagoras, \((\sqrt{n})^2+7^2=n+49\). Changing the perpendicular changes the usual sequence.

The square root spiral uses the Pythagorean theorem to create the next length. When a perpendicular of 1 unit is drawn on a hypotenuse of length \(\sqrt{1224}\), the old hypotenuse and the new perpendicular become the two perpendicular sides of a right triangle. Their squared lengths must be added.

The new hypotenuse is \(\sqrt{(\sqrt{1224})^2+1^2}=\sqrt{1224+1}=\sqrt{1225}\). Since \(35^2=1225\), this length is exactly 35, not merely a value between 34 and 35. Thus option D is correct. Options A, B and C use the wrong arithmetic or give a nonmatching interval; in particular, \(\sqrt{1223}\) is less than 35.

Since \(22^2=484\) and \(23^2=529\), and \(484<528<529\), we get \(22<\sqrt{528}<23\). Similarly, \(530\) lies between \(23^2=529\) and \(24^2=576\), so \(23<\sqrt{530}<24\). Hence, option B is correct. Option C may seem close, but \(528<529\) means that \(\sqrt{528}\) is less than 23. Exam tip: locate a square root by comparing the number with consecutive perfect squares.

In a square root spiral, the hypotenuse after \(\sqrt{1520}\) is \(\sqrt{1521}\). Since \(1521=39^2\), we get \(\sqrt{1521}=39\). \(38\) is incorrect because \(38^2=1444\), while \(40^2=1600\). Exam tip: To check whether a square root is an integer, compare the number with nearby perfect squares.

If the (k)-th hypotenuse is (\sqrt{k}), then (\sqrt{144}) is the (144)-th hypotenuse. Also, (\sqrt{144}=12).

\((\sqrt{389})^2+1^2=390\). So the previous hypotenuse for \(\sqrt{390}\) is \(\sqrt{389}\).