Square root spiral

By Pythagoras the new hypotenuse is (\sqrt{(\sqrt{13})^2+1^2}=\sqrt{14}). In a square root spiral the number increases by one.

If the previous hypotenuse is \(\sqrt{n}\) and a unit side is drawn perpendicular to it, the new hypotenuse has square \(n+1\): \((\sqrt{n})^2+1^2=n+1\). Repeatedly using the initial line breaks this sequence. Exam tip: check “previous hypotenuse.”

The next hypotenuse is (\sqrt{36}) and (\sqrt{36}=6). It lies exactly at (6), not between (5) and (6).

The square of the \(\sqrt{16}\) side is 16, and the square of the new perpendicular side is 1. By Pythagoras, the new hypotenuse has square \(16+1=17\), so it is \(\sqrt{17}\). Exam tip: each new spiral step uses a 1-unit perpendicular side.

The expression \(\sqrt{7}+1\) adds two lengths directly, but that is not how a square root spiral creates its next hypotenuse. The segment of length 1 is perpendicular to the previous hypotenuse, so the two lengths are the legs of a right triangle. The diagonal must be calculated from their squared lengths.

For the old length \(\sqrt{7}\), Pythagoras gives \(h=\sqrt{(\sqrt{7})^2+1^2}=\sqrt{7+1}=\sqrt{8}\). Thus option A gives both the correct method and result. Option D wrongly uses \(7\) as the length before squaring, even though the actual length is \(\sqrt{7}\). The other options also use invalid direct addition or subtraction.

(48) is not a perfect square but (49) is a perfect square. Therefore (\sqrt{48}) is irrational and (\sqrt{49}=7).

With (\sqrt{43}) and a (1) unit perpendicular, the new hypotenuse becomes (\sqrt{44}). The previous hypotenuse has one less number.

In the spiral, a 1-unit perpendicular leg is drawn on the previous hypotenuse \(\sqrt{n-1}\). By Pythagoras, hypotenuse² = \((n-1)+1=n\), so it becomes \(\sqrt n\). Exam tip: remember the added leg is always 1 unit.

If the new hypotenuse is (\sqrt{n+1}), the previous one is (\sqrt{n}). Therefore before (\sqrt{82}), it was (\sqrt{81}).

(\sqrt{81}=9), and after adding a (1) unit perpendicular the next hypotenuse is (\sqrt{82}). The number increases by one.

In the usual sequence (1^2) is added each time. Taking (3) units adds (9), so the sequence changes.

In a square root spiral, each new hypotenuse represents the required square root. Hence the point for \(\sqrt{10}\) is \(\sqrt{10}\) units from the origin, not 10 units. Since \(3^2<10<4^2\), its length lies between 3 and 4. Exam tip: never confuse a number with its square root.

(\sqrt{2}) becomes the previous hypotenuse side. A (1) unit perpendicular at its end gives the new hypotenuse (\sqrt{3}).

\(10^2=100\) and \(11^2=121\). Since \(100<120<121\), taking square roots gives \(10<\sqrt{120}<11\). The interval \(11<\sqrt{120}<12\) is incorrect because it would require the number inside the root to be greater than \(121\). Exam tip: Compare a number with the nearest perfect squares to locate its square root.

(\sqrt{143}) is formed from (\sqrt{142}) and a (1) unit perpendicular. The number increases by one in the next hypotenuse.

In a square root spiral, a perpendicular side of 1 unit is drawn on the previous hypotenuse. By Pythagoras, the new hypotenuse becomes \(\sqrt{n+1}\). Exam tip: remember “1-unit perpendicular” as the key construction rule.

Since \(5^2=25\) and \(6^2=36\), the number 26 lies between these squares; hence \(\sqrt{26}\) lies between 5 and 6. Option B fails because 26 is greater than 25. Exam tip: compare with consecutive perfect squares.

(\sqrt{7}) is the previous hypotenuse and the new perpendicular is (1) unit. Therefore the new hypotenuse is (\sqrt{8}).

(169) is a perfect square and (170) is not. Therefore (\sqrt{169}=13) and (\sqrt{170}) is irrational.

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