Square root spiral

(\sqrt{2}) is the first hypotenuse, so the (25)th hypotenuse is (\sqrt{25+1}=\sqrt{26}). Keep the starting point of counting clear.

((\sqrt{12})^2+1^2=13), so the result will be (\sqrt{13}). To get (\sqrt{14}), (\sqrt{13}) is needed first.

(\sqrt{144}=12) is rational, while (145) is not a perfect square. Therefore, (\sqrt{145}) is irrational.

In a square root spiral, each newly drawn perpendicular segment increases the number under the square root by 1: √40, √41, √42, and so on. The number of steps from 40 to 47 is 47 − 40 = 7. Therefore seven new perpendicular segments are required, making option C correct. Counting the endpoints or subtracting the wrong way would produce the distractor values.

The first right triangle has perpendicular sides (1) and (1). Therefore, the hypotenuse is (\sqrt{2}).

In a square-root spiral, \(OQ\) is the hypotenuse of the next right triangle, while \(OP\) is one of its sides. By Pythagoras' theorem, \(PQ^2=OQ^2-OP^2=74-73=1\). Hence, the correct answer is \(1\). The values \(73\) and \(74\) are \(OP^2\) and \(OQ^2\), respectively, not \(PQ^2\). Exam tip: the difference between the squares of consecutive hypotenuses gives the square of the new perpendicular segment.

The spiral gives the exact geometric distance of (\sqrt{6}). The same distance can be placed on the number line using a compass.

(\sqrt{200}=\sqrt{100\times2}=10\sqrt{2}). A large perfect square factor gives the simplified form quickly.

(\sqrt{2},\sqrt{3},\sqrt{4},\sqrt{5}) are (4) hypotenuses in total. Count (\sqrt{2}) as the first hypotenuse.

Since \(75=25\times 3\) and \(25\) is a perfect square, \(\sqrt{75}=\sqrt{25\times 3}=\sqrt{25}\sqrt{3}=5\sqrt{3}\). The close distractor \(3\sqrt{5}\) squares to \(45\), not \(75\). Exam tip: to simplify a surd, first identify the greatest perfect-square factor of the number.

In a square root spiral, each new hypotenuse has a radicand that is 1 more than that of the previous hypotenuse. Hence, the hypotenuse after \(\sqrt{x}\) is \(\sqrt{x+1}\). Given \(\sqrt{x+1}=\sqrt{57}\), we get \(x+1=57\), so \(x=56\). Option 57 is the radicand of the next hypotenuse, not the value of \(x\). Exam tip: radicands of consecutive hypotenuses increase by 1.

From (\sqrt{2}) to (\sqrt{9}), the inner numbers (2,3,4,5,6,7,8,9) give (8) hypotenuses. (\sqrt{4}) and (\sqrt{9}) also give integer lengths.

Since \(27=9\times 3\), and \(9\) is a perfect square, \(\sqrt{27}=\sqrt{9\times 3}=\sqrt{9}\sqrt{3}=3\sqrt{3}\). The expression \(9\sqrt{3}\) has an extra factor of 3, while \(\sqrt{9}=3\) is not equal to \(\sqrt{27}\). Exam tip: To simplify a square root, identify the greatest perfect-square factor inside the radical and take it outside.

The two given sides form a right triangle, with \\(OP=\\sqrt{18}\\) as one leg and \\(PQ=1\\) as the perpendicular leg. The segment from O to Q is the hypotenuse. By the Pythagorean theorem, the square of the hypotenuse equals the sum of the squares of the perpendicular sides. Therefore its length is \\(\\sqrt{18+1}=\\sqrt{19}\\), so option C is correct.

More explicitly, \\(OQ^2=(\\sqrt{18})^2+1^2=18+1=19\\), and hence \\(OQ=\\sqrt{19}\\). The number 19 is not a perfect square and has no square factor greater than 1, so the radical cannot be simplified further. It is not \\(\\sqrt{18}\\), because adding a nonzero perpendicular side increases the hypotenuse.

The square root of a number is not automatically irrational. A square root is rational when the number under the root is a perfect square. Since 4 is the square of the integer 2, \(\sqrt{4}=2\). The number 2 can be written as \(2/1\), so it is rational. Therefore, option A gives the correct reason. The fact that a length is shown in a square-root spiral does not change its numerical nature. A spiral may represent both rational and irrational lengths.

To check the answer, compare the choices. Option B is false because \(\sqrt{4}\) is not 4. Option C is false because 4 is composite, not prime. Option D is also false because the spiral does not produce only rational numbers; for example, \(\sqrt{2}\) is irrational. Thus the exact calculation \(\sqrt{4}=2\) proves that the value is rational, so answer A follows.

In a square root spiral, the hypotenuse of the (m)th right triangle is \(\sqrt{m+1}\). Hence, \(\sqrt{m+1}=\sqrt{46}\), so \(m+1=46\) and \(m=45\). Option 46 is incorrect because it is the square of the hypotenuse, not the triangle number. Exam tip: verify the pattern by recalling that the first triangle has hypotenuse \(\sqrt{2}\).

(\sqrt{4}=2), (\sqrt{9}=3), and (\sqrt{16}=4) are integers. Therefore, (3) integer lengths are obtained.

By Pythagoras theorem, the next hypotenuse is (\sqrt{(\text{previous hypotenuse})^2+1}). So both square and square root must be in the correct places.

If the length is (5), its square is (25), so the hypotenuse appears as (\sqrt{25}). Connect integer lengths with their square root forms.

In a square root spiral, every new hypotenuse is formed using a right triangle. Therefore, its main basis is Pythagoras theorem.