Square root spiral

Use the square-root spiral rule and convert the given whole-number hypotenuse into radical form. Since √(n+1) = 12, squaring both sides gives n + 1 = 144, so n = 143. The previous hypotenuse was therefore √n = √143. Option B is correct; √144 is the new hypotenuse itself, while √142 and √145 do not satisfy the equation.

The hypotenuses increase in order in the spiral. Just before \(\sqrt{10}\) comes \(\sqrt{9}\), and before that \(\sqrt{8}\).

Actually (13^2=169) and (14^2=196). Therefore (\sqrt{195}) lies between (13) and (14), less than (14).

In Pythagoras theorem, the square of the hypotenuse equals the sum of squares of the sides. This is the general rule.

Since \(196=14^2\), \(\sqrt{196}=14\), which is rational. The numbers 198, 200 and 202 are not perfect squares, so their square roots are irrational. Exam tip: check for a perfect square first.

\(\sqrt{n}\) is rational only when \(n\) is a perfect square. For example, \(\sqrt{49}=7\), so that spiral segment has rational length. Being prime or odd is not sufficient. Exam tip: check for perfect squares first.

Let the previous hypotenuse be \(\sqrt{n}\). With a new perpendicular side of 1, Pythagoras gives \(\sqrt{(\sqrt{n})^2+1^2}=\sqrt{n+1}\). It is not obtained by simply adding 1 to the length. Exam tip: add the squares of perpendicular sides.

(\sqrt{224}) is formed from (\sqrt{223}). Since (14^2<224<15^2), it lies between (14) and (15).

Since \(4^2=16\) and \(5^2=25\), \(\sqrt{18}\) must lie between 4 and 5. Its value is about 4.24, while \(4.5^2=20.25\). Exam tip: compare the nearest perfect squares first.

The same length to be marked is taken in the compass. Drawing an arc from the origin gives the correct position.

Each new right triangle in the square root spiral has a perpendicular side of 1 unit. \(\sqrt{9^2+1^2}=\sqrt{10}\), so a 10-unit base is not needed. Exam tip: apply Pythagoras' theorem.

(\sqrt{4}) is the previous hypotenuse and (1) is the new perpendicular. Pythagoras gives the new hypotenuse (\sqrt{5}).

\(20^2=400\) and \(21^2=441\). Since \(400<440<441\), taking square roots gives \(20<\sqrt{440}<21\). Although it is close to \(21\), \(\sqrt{440}\) is less than \(21\) because \(440<441\). Exam tip: To locate a square root, compare the number with consecutive perfect squares.

If the (k)-th hypotenuse is (\sqrt{k}), then (\sqrt{25}) is the (25)-th hypotenuse. Also, (\sqrt{25}=5).

Apply Pythagoras to the new hypotenuse: \(h^2=14+1=15\), so \(h=\sqrt{15}\). Directly adding 1 to \(\sqrt{14}\) is invalid. Exam tip: add squares, not lengths.

The correct basis of a square root spiral is a right triangle and Pythagoras theorem. Directly adding lengths is wrong.

At each step, a unit segment is added perpendicular to the previous hypotenuse, creating a right triangle. By Pythagoras,

g(

gsqrt{n})^2+1^2=n+1, so the next hypotenuse is

gsqrt{n+1}. Exam tip: look for “perpendicular” and “unit length.”

Because (24^2=576) and (25^2=625). The number (624) lies between them, so (\sqrt{624}) lies between (24) and (25).

The square root of a positive integer is a whole number only when it is a perfect square. If it is not a perfect square, the square root is irrational.

Pythagoras theorem uses the sum of squares. Therefore (\sqrt{20}) and (1) form hypotenuse (\sqrt{21}).