Square root spiral

Since \(3^2=9\) and \(4^2=16\), we have \(9<10<16\), so \(3<\sqrt{10}<4\). The value inside the root is not compared directly with 4. Exam tip: bracket a square root using the nearest perfect squares.

The statement is correct. With previous hypotenuse \(\sqrt{7}\) and a new perpendicular of 1, Pythagoras gives hypotenuse² = 7 + 1 = 8, so the new hypotenuse is \(\sqrt{8}\). Exam tip: add squares, not the lengths themselves.

With a (1) unit perpendicular and a (90^\circ) angle, Pythagoras theorem applies correctly. This forms successive square roots.

In a standard square root spiral, each new perpendicular is 1 cm, so \(h^2\) increases by 1 at every step. With a 2 cm side, \(h^2\) increases by 4 instead. Exam tip: always check the unit perpendicular.

At each step, a new unit segment is drawn perpendicular to the previous hypotenuse. If that hypotenuse is \(\sqrt{n}\), the next one becomes \(\sqrt{n+1}\). In exams, identify the newly formed outer right triangle.

Each new right triangle in the spiral is formed by adding a side of length 1. By Pythagoras’ theorem, successive hypotenuse squares are 2, 3, 4, …, so A is correct. Exam tip: check the sequence of squared lengths first.

Since 17 is not a perfect square, \(\sqrt{17}\) cannot be written as a ratio of two integers. Hence, its point on the spiral represents an irrational number. Exam tip: only square roots of perfect squares are integers.

Adding a (1) unit perpendicular to (\sqrt{254}) gives (\sqrt{255}). The previous hypotenuse has one less number.

The rule of the square root spiral is based on Pythagoras theorem and a (1) unit perpendicular. This forms successive square roots.

The new hypotenuse is (\sqrt{120+1}=\sqrt{121}), and (\sqrt{121}=11). When a perfect square appears, write its exact value.

The correct hypotenuse is (\sqrt{(\sqrt{18})^2+1^2}=\sqrt{19}). In a square root spiral, the number increases by (1) at each step.

The statement is incorrect. Adding a perpendicular side of length 1 gives successive hypotenuses \(\sqrt{n}\); for example, \(\sqrt{3}\) follows \(\sqrt{2}\). Thus, non-perfect-square roots also occur. Exam tip: track the radicand step by step.

If the new hypotenuse is (\sqrt{n+1}), the previous hypotenuse is (\sqrt{n}). Therefore before (\sqrt{35}), it is (\sqrt{34}).

\(6^2=36\) and \(7^2=49\). Since \(36<45<49\), we get \(6<\sqrt{45}<7\). Therefore, the length \(\sqrt{45}\) in the square root spiral lies between \(6\) and \(7\) on the number line. It cannot lie between \(5\) and \(6\), because numbers in that interval have squares between \(25\) and \(36\). Exam tip: compare the radicand with the nearest perfect squares to locate a square root.

The previous hypotenuse is \(\sqrt{12}\) and the new perpendicular side is 1 unit. By Pythagoras, its square is \((\sqrt{12})^2+1^2=12+1=13\), so the new hypotenuse is \(\sqrt{13}\). Exam tip: add squares of perpendicular sides.

A 1-unit side is drawn perpendicular to the previous hypotenuse, forming a right triangle. By Pythagoras, new hypotenuse² = previous hypotenuse² + 1, giving \(\sqrt{2},\sqrt{3}\), etc. Exam tip: look for “perpendicular,” not parallel.

Since \(8^2=64<72<81=9^2\), \(\sqrt{72}\) lies between \(8\) and \(9\). Also, \(81=9^2\), so \(\sqrt{81}=9\), which is a whole number. Therefore, option B is correct. Option D is incorrect because \(\sqrt{81}\) is not irrational. Exam tip: To locate a square root, compare the number with the squares of nearby whole numbers.

(\sqrt{120}) is formed from (\sqrt{119}) and a (1) unit perpendicular. The previous hypotenuse number is always (1) less.

Since \(10^2=100\) and \(11^2=121\), and \(100<120<121\), we get \(10<\sqrt{120}<11\). Therefore, \(\sqrt{120}\) lies between 10 and 11 on the number line. For it to lie between 9 and 10, its radicand would need to be between 81 and 100. Exam tip: compare the number with the nearest perfect squares to locate a square root.

In the usual spiral (1^2) is added every time. Taking (4) units adds (16), so the sequence changes.