Square root spiral

By Pythagoras theorem ((\sqrt{n})^2+1^2=n+1). Therefore the new hypotenuse becomes (\sqrt{n+1}).

Taking the hypotenuse length in the compass, an arc is drawn from the origin. This gives the correct point on the number line.

In a square root spiral, each new right triangle uses the previous hypotenuse and a new perpendicular side of 1 unit. By Pythagoras, the new hypotenuse satisfies \(h^2=a^2+1^2\). Exam tip: identify the added 1-unit perpendicular.

Since \(36=6^2\), \(\sqrt{36}=6\), which is a whole number. However, \(37\) is not a perfect square, so \(\sqrt{37}\) is irrational. Option D is incorrect because only \(36\), not \(37\), is a perfect square. Exam tip: The square root of a number is a whole number only when the number is a perfect square.

In a square root spiral, a unit-length side is drawn perpendicular to the previous hypotenuse. By Pythagoras’ theorem, the new hypotenuse becomes \(\sqrt{2},\sqrt{3},\sqrt{4}\), and so on. Exam tip: remember “perpendicular unit side.”

In the usual spiral (1^2) is added each time to give the next square root. Taking (2) units will add (4).

In a square root spiral, the successive hypotenuses are \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on; the number under the square root increases by 1 at each step. Therefore, the hypotenuse after \(\sqrt{99}\) is \(\sqrt{100}\). \(\sqrt{98}\) is the previous hypotenuse, while \(\sqrt{198}\) is not the next term in this sequence. Exam tip: the squares of successive spiral hypotenuses are consecutive natural numbers.

A square root spiral is a chain of right triangles. The previous hypotenuse and a (1) unit perpendicular form the new triangle.

Each new triangle is formed by drawing a perpendicular side of length 1 on the previous hypotenuse \(\sqrt{n-1}\). By Pythagoras, \((\sqrt{n-1})^2+1^2=n\), so its hypotenuse is \(\sqrt{n}\). The 1-unit segment is a leg, not the hypotenuse. Exam tip: every new hypotenuse represents the next square root.

The previous hypotenuse (\sqrt{2}) becomes one side and a (1) unit perpendicular is added. The new hypotenuse is (\sqrt{3}).

Since \(5^2=25\) and \(6^2=36\), and \(25<32<36\), we get \(5<\sqrt{32}<6\). Therefore, \(\sqrt{32}\) lies between 5 and 6 on the number line. It cannot lie between 4 and 5, because that interval corresponds to numbers between 16 and 25. Exam tip: compare the number with the nearest perfect squares to locate its square root.

Adding a (1) unit perpendicular to (\sqrt{47}) forms (\sqrt{48}). In the previous step the number is (1) less.

In the usual sequence of the square root spiral, (\sqrt{8}) comes after (\sqrt{7}). The new perpendicular is always (1) unit.

\(20\) is not a perfect square. Since \(20=4\times5\), \(\sqrt{20}=2\sqrt{5}\). As \(\sqrt{5}\) is irrational, \(\sqrt{20}\) is also irrational. Being even does not make \(20\) or its square root zero. Exam tip: The square root of an integer is rational only when the integer is a perfect square.

In each new triangle, the previous hypotenuse becomes one leg and a perpendicular unit-length leg is added. If the old hypotenuse is \(\sqrt{n}\), Pythagoras gives the new one as \(\sqrt{n+1}\). Two unit legs produce only \(\sqrt{2}\). Exam tip: identify the fixed unit leg first.

In the spiral, the previous hypotenuse is \(\sqrt{n}\). A perpendicular unit side gives the next hypotenuse \(\sqrt{n+1}\), since \((\sqrt{n})^2+1^2=n+1\). Exam tip: check that each new unit segment is perpendicular to the previous hypotenuse.

The new hypotenuse is \(\sqrt{6}\), since \(h^2=(\sqrt{5})^2+1^2=5+1=6\). It is not \(\sqrt{10}\), as the lengths are not added directly. Exam tip: apply Pythagoras’ theorem at every new step.

The spiral does not obtain a new length by simply adding the two side lengths. Instead, each new figure is a right triangle, so the Pythagorean theorem must be used. The square of the new hypotenuse equals the sum of the squares of the old hypotenuse and the new unit perpendicular. This explains why the radical changes from 2 to 3.

For the given construction, the two perpendicular sides have lengths
sqrt{2}
and 1. Therefore the new hypotenuse
h
satisfies
h^2=(
sqrt{2}
)^2+1^2=2+1=3
. Since a length is positive,
h=
sqrt{3}
. The expression
sqrt{2}+1
is not equal to
sqrt{3}
, and the other numerical statements do not represent the theorem. Hence option A is correct.

In a square root spiral, a perpendicular side of length 1 unit is drawn from an end of the previous hypotenuse. By Pythagoras, the new hypotenuse becomes \(\sqrt{n+1}\). Exam tip: the added side is always 1 unit.

Since \(25<27<36\), we have \(5^2<27<6^2\). Taking square roots gives \(5<\sqrt{27}<6\), so it must be placed between \(5\) and \(6\) on the number line. The interval \(4\) and \(5\) is incorrect because \(\sqrt{25}=5\) and \(27\) is greater than \(25\). Exam tip: To locate \(\sqrt{n}\), compare \(n\) with the nearest perfect squares.