Square root spiral

(\sqrt{49}=7), which is a natural number. Square roots of perfect squares can be natural numbers.

In the next step, the inner number increases by (1), so (\sqrt{81}) is formed. The spiral does not skip numbers in order.

In the standard square-root spiral construction, \(PQ\) is perpendicular to \(OP\). By Pythagoras’ theorem, \(OQ^2=OP^2+PQ^2=(\sqrt{a})^2+1^2=a+1\). Hence, \(OQ=\sqrt{a+1}\). The expression \(a+1\) is the value of \(OQ^2\), not the length of the hypotenuse itself. Exam tip: add the squares of the perpendicular sides first, then take the square root.

In order, the inner number decreases by (1) before and increases by (1) after. So (\sqrt{98}) comes before (\sqrt{99}), and (\sqrt{100}) comes after it.

\(121\) is a perfect square because \(11 \times 11 = 121\). Therefore, \(\sqrt{121}=11\), so its actual length in the spiral is 11 units. \(121\) is the number under the square root, not its square root. Exam tip: Memorising squares from 1 to 15 helps solve such questions quickly.

At each new step, the inner number increases by (1), so (8+5=13). The final hypotenuse is (\sqrt{13}).

In a square root spiral, each new step increases the number under the square root of the hypotenuse by 1. Since 7 new steps lead to \(\sqrt{29}\), the starting number was \(29-7=22\). Hence, the starting hypotenuse was \(\sqrt{22}\). If it had been \(\sqrt{21}\), seven steps would give \(\sqrt{28}\), not \(\sqrt{29}\). Exam tip: Add or subtract the number of steps from the radicand, not from the square root itself.

Since \(54=9\times6\), and \(9\) is the greatest perfect-square factor of 54, \(\sqrt{54}=\sqrt{9\times6}=\sqrt9\times\sqrt6=3\sqrt6\). The expression \(2\sqrt{27}\) is neither in simplest form nor equal to \(\sqrt{54}\). Exam tip: first identify the greatest perfect-square factor inside the radical.

The main construction depends on drawing a (1) unit perpendicular to the previous hypotenuse. This creates the next right triangle.

At each new step, the perpendicular is drawn on the previous hypotenuse, so the direction gradually changes. The spiral is formed by this turning.

(\sqrt{16}) is certain only when the new segment is perpendicular to the previous hypotenuse. Without a right angle, Pythagoras theorem cannot be applied.

In a square root spiral, each new step represents the next number: \(\sqrt{1}, \sqrt{2}, \sqrt{3}, \ldots\). From \(\sqrt{3}\) to \(\sqrt{7}\), the radicand changes from 3 to 7, so the number of new steps is \(7-3=4\). The four steps correspond to \(\sqrt{4}, \sqrt{5}, \sqrt{6}\), and \(\sqrt{7}\). Exam tip: subtract the starting radicand from the ending radicand in such questions.

Since \(90=9\times10\) and \(9\) is a perfect square, \(\sqrt{90}=\sqrt{9\times10}=\sqrt9\sqrt{10}=3\sqrt{10}\). \(9\sqrt{10}\) is incorrect because \(\sqrt9=3\), not 9. Exam tip: identify the greatest perfect-square factor when simplifying a surd.

\((\sqrt{26})^2=26\) and \(1^2=1\), so the sum is (27). This is the step to form \(\sqrt{27}\).

The next hypotenuse is (\sqrt{101}), which is not a simple integer. Even after (\sqrt{100}=10), the sequence continues with (\sqrt{101}).

The spiral gives an exact distance that can be transferred to the number line with a compass. It is useful for showing irrational numbers.

In a square root spiral, the number under the radical increases by 1 at each successive step. Going 3 steps back from \(\sqrt{65}\) gives \(65-3=62\). Hence, the required hypotenuse is \(\sqrt{62}\). \(\sqrt{63}\) is only 2 steps before the final hypotenuse. Exam tip: subtract the number of backward steps from the radicand.

A square-root spiral is built by repeatedly drawing a new segment perpendicular to the previous segment. In the standard construction, each newly drawn perpendicular segment has a fixed length of 1 unit. The initial segment also has length 1 unit, and these two perpendicular unit segments form the first right triangle. Its hypotenuse is therefore
(1^2+1^2)=\sqrt{2}
, which begins the sequence of square roots.

Thus, the statements saying that the initial segment is 1 unit, every new perpendicular segment is 1 unit, and the first triangle uses two segments of length 1 are consistent with the standard construction. Choosing a new perpendicular segment of length 2 changes the Pythagorean calculation and produces a different construction, not the usual square-root spiral. Therefore option C is the incorrect fixed length. The supplied answer and explanation are accurate.

(\sqrt{120}=\sqrt{4\times30}=2\sqrt{30}) is simplest, while (4\sqrt{15}) squares to (240). Check each option carefully.

Since \(120=4\times30\), and \(4\) is a perfect square, \(\sqrt{120}=\sqrt{4\times30}=2\sqrt{30}\). As 30 has no perfect-square factor greater than 1, this is the simplest form. The close distractor \(4\sqrt{15}\) is incorrect because its square is \(16\times15=240\), not 120. Exam tip: identify the greatest perfect-square factor inside the radical before simplifying.