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.

The relevant concept is the classification of square roots. Since 144 is a perfect square, √144 = 12, which is an integer and therefore also a rational number. In contrast, 145 lies between the consecutive perfect squares 144 and 169, so it is not a perfect square. The square root of a positive integer that is not a perfect square is irrational; hence √145 is irrational. Therefore option B is correct. Option A reverses the classifications. Option C incorrectly treats √145 as rational, and option D is false because √145 is not an integer.

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.

Direct answer: Option A, \(10\sqrt{2}\). To simplify a square root, look for a perfect-square factor inside the radicand. Since \(200=100\times2\), \(\sqrt{200}=\sqrt{100\times2}=\sqrt{100}\sqrt{2}=10\sqrt{2}\). Option A is therefore correct. Option B, \(20\sqrt{2}\), has square \((20\sqrt{2})^2=800\), so it is twice as large as the required value. Option C, \(100\sqrt{2}\), is also much too large; it treats 100 as though it were the square root of 100. Option D, \(2\sqrt{100}=2\times10=20\), is not equal to \(10\sqrt{2}\); it incorrectly separates the factors because the remaining factor is 2, not a perfect square. The valid rule is \(\sqrt{ab}=\sqrt a\sqrt b\) for non-negative factors, and we take the largest useful perfect square. Memory cue: split the number into a perfect square times the leftover factor.

(\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.

By Pythagoras, the hypotenuse is ( \sqrt{1^2+1^2}=\sqrt{2} ). In exams, connect the first triangle with ( \sqrt{2} ).

The new radius is ( \sqrt{(\sqrt{8})^2+1^2}=\sqrt{9}=3 ). Remember that each new step adds (1) inside the square root.

For ( \sqrt{5} ), a unit segment (1) is drawn perpendicular to ( \sqrt{4} ). Perpendicularity is the key construction clue.

The direct answer is B: the next radius cannot be surely taken as \\(\\sqrt{8}\\). In the square-root spiral, after a radius \\(\\sqrt{7}\\), a unit segment is drawn perpendicular to it. Pythagoras then gives \\(\\sqrt{(\\sqrt{7})^2+1^2}=\\sqrt{7+1}=\\sqrt{8}\\). The perpendicular condition creates the right triangle; without it, the Pythagorean calculation does not apply, so the length is not guaranteed. A is wrong because the required right angle is missing. B is correct. C is wrong because the radius is not automatically the number 8. D is wrong because an incorrect construction does not necessarily close the spiral. Memory cue: square-root spiral means unit length plus a right angle.

From ( \sqrt{2} ) to ( \sqrt{10} ), the inside number increases (8) times. Each increase needs one new unit segment.

The next triangle has perpendicular sides ( \sqrt{6} ) and (1). Its area is ( \frac{1}{2}\times \sqrt{6}\times 1=\frac{\sqrt{6}}{2} ).

Since (13) is not a perfect square, ( \sqrt{13} ) is irrational. The square root spiral shows such square roots geometrically.