Square root spiral

Each added triangle must be right-angled. Drawing a unit side perpendicular to the previous hypotenuse gives the next hypotenuse using Pythagoras’ theorem. A parallel side does not form a right angle. Exam tip: mark the new unit side perpendicular first.

The first right triangle in a square root spiral has perpendicular sides of 1 unit and 1 unit. By Pythagoras, hypotenuse² = 1² + 1² = 2, so the hypotenuse is \(\sqrt{2}\). A square of side 2 has diagonal \(2\sqrt{2}\), not \(\sqrt{2}\). Exam tip: identify the triangle’s legs first.

In each new triangle, the previous hypotenuse is one leg and a unit segment is drawn perpendicular to it. By Pythagoras, the square of the new hypotenuse increases by 1. Exam tip: each added segment is perpendicular to the previous hypotenuse.

Since 7 is not a perfect square, \(\sqrt{7}\) is irrational and is shown as a distinct length on the spiral. In contrast, \(\sqrt{9}=3\), \(\sqrt{16}=4\), and \(\sqrt{25}=5\) are rational. Exam tip: square roots of perfect squares are integers.

The first hypotenuse is \(\sqrt{2}\). Adding a new 1-unit side gives hypotenuse squared as \(2+1=3\), so the next hypotenuse is \(\sqrt{3}\). Exam tip: each added triangle is right-angled.

By Pythagoras’ theorem, hypotenuse² = 1² + 1² = 2, so the hypotenuse is \(\sqrt{2}\) units. Option 2 is only the sum of the sides, not the hypotenuse. Exam tip: add squares, then take the square root.

In a square root spiral, each new right triangle uses the previous hypotenuse and a new side of 1 unit. Since \(\sqrt{16}^2+1^2=16+1=17\), the new hypotenuse is \(\sqrt{17}\). A side of 4 gives \(16+16=32\), not 17. Exam tip: check the sum of squares using Pythagoras' theorem.

Adding a (1) unit perpendicular to (\sqrt{56}) forms (\sqrt{57}). The previous hypotenuse number is always (1) less.

The square-root spiral advances by one under the radical at every stage. The construction uses the current hypotenuse and a new perpendicular segment of length 1. If the current hypotenuse is
sqrt{n}
, then the next hypotenuse has square length
n+1
, so its length is
sqrt{n+1}
. This is a direct application of the Pythagorean theorem.

Here the current value is
sqrt{43}
. Adding the square of the new unit side gives
(
sqrt{43}
)^2+1^2=43+1=44
. Taking the positive square root, as lengths are positive, gives
sqrt{44}
. Therefore option C is correct.
sqrt{42}
is the preceding stage, while
sqrt{43}
is the existing hypotenuse, not the new one. There is no doubling in this process.

A ruler is useful for taking the (1) unit length correctly. Wrong measurement can make the spiral lengths incorrect.

The previous hypotenuse is (\sqrt{11}) and the new perpendicular is (1) unit. So the new hypotenuse will be (\sqrt{12}).

Adding a (1) unit perpendicular to (\sqrt{16}) forms the new hypotenuse (\sqrt{17}). The previous hypotenuse has one less number.

Option A is correct. Since \(3^2=9\) and \(4^2=16\), we get \(9<10<16\), so \(3<\sqrt{10}<4\). Thus, the \(\sqrt{10}\) point lies between 3 and 4. Exam tip: compare with nearby perfect squares first.

In a square root spiral, one leg of the new right triangle is the hypotenuse of the previous triangle, while the other leg is 1 unit. By Pythagoras’ theorem, the new hypotenuse represents the next square root. Exam tip: identify the previous hypotenuse as the shared side.

In a square root spiral, the previous hypotenuse becomes one side and a unit segment is drawn perpendicular to it. By Pythagoras, \\(\sqrt{n}^2+1^2=n+1\\), so the new hypotenuse is \\(\sqrt{n+1}\\). Exam tip: remember “perpendicular unit side.”

The square root spiral uses right triangles to construct successive lengths. If the existing hypotenuse is (\sqrt{3}) and a perpendicular side of length 1 is added, the new hypotenuse is found by the Pythagorean theorem. It is not obtained by simply adding 1 to the old length. Therefore, the mistake to avoid is assuming that \sqrt{3}+1=\sqrt{4}.

Indeed, the square of the new hypotenuse is (\sqrt{3})^2+1^2=3+1=4, so its length is \sqrt{4}=2. Thus option A identifies the common error. Option B states the correct squared-length relation, while making a right angle and drawing the unit perpendicular are necessary construction steps.

If the new hypotenuse is (\sqrt{n+1}), the previous one is (\sqrt{n}). Therefore before (\sqrt{31}), it was (\sqrt{30}).

Since \(49=7^2\), \(\sqrt{49}=7\), which is a whole number. However, \(50\) is not a perfect square; it lies between \(49\) and \(64\), so \(\sqrt{50}\) is not a whole number and is irrational. Option C is incorrect because only \(\sqrt{49}\) is a whole number. Exam tip: A square root is a whole number only when the number is a perfect square.

In a square root spiral, each new triangle must be right-angled. Drawing the 1-unit segment at 90° to the previous hypotenuse allows Pythagoras’ theorem to be used, giving successive hypotenuse lengths such as \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. A 1-unit side alone does not make the triangle equilateral; the right angle is essential. Exam tip: For successive square-root lengths obtained as hypotenuses, look for a right triangle and use Pythagoras’ theorem.

The previous hypotenuse \(\sqrt{n-1}\) with a unit perpendicular side gives \(\sqrt{(n-1)+1}=\sqrt n\) by Pythagoras. The sides are not parallel. Tip: mark the right angle.