Square root spiral

The same hypotenuse length must be taken in the compass. Drawing an arc from the origin gives the correct point.

The governing concept is the Pythagorean theorem in the construction of a square root spiral. If the existing hypotenuse has length √n and a new perpendicular side of length 1 is added, the new hypotenuse h satisfies h²=(√n)²+1²=n+1. Thus h=√((√n)²+1²)=√(n+1), which is the complete and correct form in option A. The expression √n+1 means ordinary addition outside the radical and is generally not equal to √(n+1). Option B has no valid Pythagorean derivation, option C changes addition to subtraction, and option D incorrectly treats n rather than √n as the existing length. Therefore A correctly repairs the shortcut.

First the exact square root length is constructed. Then the same length is marked on the number line using a compass.

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.

The direct answer is B: \(\sqrt{48}\) is irrational and \(\sqrt{49}=7\). First check whether each number inside the square root is a perfect square. The nearby perfect squares are \(36=6^2\) and \(49=7^2\). Since 48 is not a perfect square, \(\sqrt{48}\) cannot simplify to a whole number; in fact it is irrational. Since 49 is a perfect square, \(\sqrt{49}=7\), which is a whole and rational number. Option A reverses both facts, so it is wrong. Option B states both correct facts, so it is right. Option C is wrong because \(\sqrt{48}\) is not a whole number. Option D is wrong because only \(\sqrt{49}\) equals 7; \(\sqrt{48}\) is less than 7 because 48 is less than 49, but it is not equal to 7. In the spiral, both lengths can occur as geometric lengths, but their number types differ. Memory cue: test perfect-square status first; a perfect square has a whole-number root, while a non-perfect square has an irrational root.

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.

The governing concept is the Pythagorean theorem together with the special one-unit rule used in the standard square-root spiral. If the previous hypotenuse is √n and the added perpendicular has length 1, then the new squared hypotenuse is n+1. However, if the perpendicular has length 3, its square is 3²=9, so the new hypotenuse h satisfies h²=(√n)²+3²=n+9. The sequence therefore advances through √n, √(n+9), √(n+18), and so on, rather than through consecutive roots √n, √(n+1), √(n+2). Hence option A is correct. A right angle can still be constructed, the hypotenuse is not fixed at 1, and the Pythagorean theorem remains valid.

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.

The governing construction rule is that a one-unit perpendicular changes a previous hypotenuse √n into a new hypotenuse √(n+1). This follows from the Pythagorean theorem: h²=(√n)²+1²=n+1. To obtain √143 as the new hypotenuse, we need n+1=143, so n=142. Therefore the perpendicular must be drawn on the previous hypotenuse √142, and the resulting hypotenuse will be √143. Option B is correct. Starting with √141 would produce √142, so it is one step too early. Choosing √143 confuses the required new hypotenuse with the previous one, while √144 would produce √145 and is one step too late. The direction of the construction is essential.

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.

The governing concept is the Pythagorean theorem applied to consecutive steps of the standard square-root spiral. To construct √8, the previous hypotenuse must be √7, and the new perpendicular must have length 1. If the new hypotenuse is h, then h²=(√7)²+1²=7+1=8, so h=√8. Therefore option A gives the correct equation. Option B starts with √8 as though it were the previous hypotenuse; its left side equals 8+1=9, not 8. Option C uses a 2-unit perpendicular, producing 6+4=10 rather than 8. Option D is not the Pythagorean relation because the theorem combines squares of the legs, not their lengths by ordinary addition.

(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.

In a square root spiral, each new hypotenuse represents the next square root. Hence, after \(\sqrt{55}\), the next hypotenuse is \(\sqrt{56}\). Since \(7^2=49\) and \(8^2=64\), and \(49<56<64\), \(\sqrt{56}\) lies between \(7\) and \(8\). \(\sqrt{54}\) is the preceding value, while \(\sqrt{57}\) comes after it. Exam tip: compare a number with its nearest perfect squares to locate its square root.

In a square root spiral, a perpendicular side of length \(1\) is drawn at an endpoint of the previous hypotenuse. The new hypotenuse then becomes \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. \(\sqrt{2}\) is an early hypotenuse, not the added side. Exam tip: the newly added side is always \(1\).