Square root spiral

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.

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.

The governing concepts are the classification of square roots and comparison with consecutive perfect squares. The number 12 is not a perfect square, so √12 is not a whole number. In fact, the square root of a positive integer that is not a perfect square is irrational, so √12 is irrational. To locate it on the number line, compare 12 with 3² and 4²: 3²=9 and 4²=16. Since 9<12<16, taking positive square roots gives 3<√12<4. Therefore option B is correct. It cannot be at 3 because 3²=9, and it cannot be at 4 because 4²=16. Option C places it in the wrong interval, while A and D wrongly call it a whole number.

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.

The governing concept is the geometric construction rule of the square-root spiral, not ordinary addition of lengths. At each stage, the previous hypotenuse is retained as one leg of a new right triangle, and a perpendicular segment of length 1 unit is drawn as the second leg. Pythagoras then determines the new hypotenuse. Starting with √1, we get (√1)²+1²=1+1=2, so the new hypotenuse is √2. Starting with √2, we get (√2)²+1²=2+1=3, so the next one is √3. Thus option A describes the common operation. Option D is a tempting but incorrect arithmetic shortcut: √1+1 is not √2 and √2+1 is not √3. The old hypotenuse is neither doubled nor erased.

\(8^2=64\) 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.

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.