Square root spiral

In a square root spiral, each new hypotenuse represents the next square root. Hence, after \(\sqrt{55}\), the next hypotenuse is \(\sqrt{56}\). Since \(7^2=49\) and \(8^2=64\), and \(49<56<64\), \(\sqrt{56}\) lies between \(7\) and \(8\). \(\sqrt{54}\) is the preceding value, while \(\sqrt{57}\) comes after it. Exam tip: compare a number with its nearest perfect squares to locate its square root.

A set square can be used to make a right angle. If the right angle is correct, Pythagoras gives the correct hypotenuse.

In a square root spiral, a perpendicular side of length \(1\) is drawn at an endpoint of the previous hypotenuse. The new hypotenuse then becomes \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. \(\sqrt{2}\) is an early hypotenuse, not the added side. Exam tip: the newly added side is always \(1\).

In the next right triangle, the previous hypotenuse \(\sqrt n\) is one leg and the new leg is 1. By Pythagoras, \((\sqrt n)^2+1^2=n+1\), so the hypotenuse is \(\sqrt{n+1}\). Option B wrongly adds outside the square root. Exam tip: add squares of the legs first.

Since \(9^2=81\) and \(10^2=100\), and \(81<96<100\), we get \(9<\sqrt{96}<10\). The interval from 8 to 9 is incorrect because its corresponding squares lie from \(64\) to \(81\). Exam tip: To locate a square root, compare the number with the nearest smaller and larger perfect squares.

In the spiral, the previous hypotenuse becomes one leg and a perpendicular unit leg is added. If it is \(\sqrt{n}\), then the new hypotenuse is \(\sqrt{n+1}\). Exam tip: apply Pythagoras’ theorem at each step.

A square root spiral is built step by step with right triangles. Typically, one side has a known length and a perpendicular segment of 1 unit is added. The Pythagorean theorem then gives the next hypotenuse, for example \(\sqrt{1^2+1^2}=\sqrt{2}\), and later constructions continue in the same way. The accuracy of every new length depends on the accuracy of the sides used before it.

If the 1-unit perpendicular is not measured correctly, the triangle does not have the intended dimensions. Its hypotenuse will therefore not have the intended square-root length, and every later triangle based on that length may also be inaccurate. Thus option A is correct. The measurement error does not change the Pythagorean theorem or remove the need for a right angle; it only makes the construction’s lengths incorrect.

(28) is not a perfect square and (5^2<28<6^2). Therefore (\sqrt{28}) is irrational and lies between (5) and (6).

In a right triangle the square of the hypotenuse equals the sum of squares of the sides. Therefore the hypotenuse is (\sqrt{2}).

\(10^2=100\) and \(11^2=121\). Since \(100<108<121\), we get \(10<\sqrt{108}<11\). Therefore, \(\sqrt{108}\) lies between 10 and 11 on the number line. It cannot lie between 9 and 10, because numbers in that interval have squares less than 100. Exam tip: Compare the number with consecutive perfect squares to find the interval of its square root.

Since \(100=10^2\), \(\sqrt{100}=10\). Also, \(10^2=100<101<121=11^2\), so \(\sqrt{101}\) lies between \(10\) and \(11\) and is not a whole number. Option C may seem close, but \(\sqrt{101}=11\) would require \(101=121\). Exam tip: To locate a square root, compare the number with nearby perfect squares.

The spiral advances from \(\sqrt{n}\) to \(\sqrt{n+1}\) by adding a perpendicular side of length 1. Therefore, to construct \(\sqrt{59}\), the preceding hypotenuse must be \(\sqrt{58}\). The numerical position is found by comparing 59 with nearby perfect squares: \(7^2=49\) and \(8^2=64\).

Since \(49<59<64\), taking positive square roots gives \(7<\sqrt{59}<8\). Thus the correct pair is \(\sqrt{58}\) and the interval between 7 and 8, as stated in option A. \(\sqrt{57}\) would lead to \(\sqrt{58}\), while \(\sqrt{60}\) is a later length. The interval 8 to 9 is also too high.

In a square root spiral the hypotenuses are formed in order. Skipping a step is not correct in the usual construction.

We have \(12^2=144\) and \(13^2=169\). Since \(145\) is greater than \(144\) but less than \(169\), \(12<\sqrt{145}<13\). Hence, \(\sqrt{145}\) lies between 12 and 13. The interval between 11 and 12 is incorrect because numbers in that interval have squares less than 144. Exam tip: Compare the number with the nearest perfect squares to locate its square root.

Each new unit segment is drawn perpendicular to the previous hypotenuse at its outer endpoint. If the earlier hypotenuse is \(\sqrt n\), the new one is \(\sqrt{(\sqrt n)^2+1^2}=\sqrt{n+1}\). Drawing repeatedly from the initial point breaks the spiral. Exam tip: check the right-angle mark.

(12) is not a perfect square and (3^2<12<4^2). Therefore it is irrational and lies between (3) and (4).

For the new right triangle, hypotenuse² = \((\sqrt{8})^2+1^2=8+1=9\), so the hypotenuse is \(\sqrt{9}\). \(\sqrt{10}\) would follow \(\sqrt{9}\), not \(\sqrt{8}\). Exam tip: add 1 at each spiral step.

At each step the previous hypotenuse becomes one side and (1) unit perpendicular becomes the other side. This gives the next hypotenuse.

\(8^2=64\) and \(9^2=81\). Since \(64<76<81\), we get \(8<\sqrt{76}<9\). Hence, the new hypotenuse in the square root spiral lies between 8 and 9. A close distractor may seem tempting because 76 is near 81, but it is still less than 9 when square-rooted. Exam tip: Compare the number with nearby perfect squares to locate its square root.

In the square-root spiral, every new triangle is right-angled. The previous hypotenuse is used as one leg, a perpendicular segment of length 1 is added, and the new hypotenuse is calculated with Pythagoras. Thus, if the old length is \(\sqrt{n}\), the new length is \(\sqrt{n+1}\), not a simple arithmetic sum.

Directly adding 1 to the previous hypotenuse is therefore incorrect. For example, from \(\sqrt{8}\), the next length is \(\sqrt{8+1}=\sqrt{9}\), not \(\sqrt{8}+1\). Options A, B, and C describe essential parts of the construction: making a right angle, using a unit perpendicular, and reusing the previous hypotenuse. Hence option D is the most incorrect.