Square root spiral

Correct answer: A. Take the (\sqrt{2}) hypotenuse length in a compass and draw an arc from the origin

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

Correct answer: A. The correct form is √((√n)² + 1²) = √(n+1)

Explanation: 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.

Correct answer: A. Construct the square root length using right triangles and transfer it to the number line with a compass

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

Correct answer: A. (\sqrt{13+1})

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

Correct answer: A. The new side of length 1 must be perpendicular to the immediately preceding hypotenuse.

Explanation: 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.”

Correct answer: B. Between (6) and (7)

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

Correct answer: A. The new hypotenuse will have length \(\sqrt{17}\).

Explanation: 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.

Correct answer: A. (\sqrt{(\sqrt{7})^2+1^2}=\sqrt{8})

Explanation: 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.

Correct answer: B. (\sqrt{48}) is irrational and (\sqrt{49}=7)

Explanation: 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.

Correct answer: B. Draw a (1) unit perpendicular on (\sqrt{43})

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

Correct answer: A. One leg is 1 unit and the other perpendicular leg is \(\sqrt{n-1}\).

Explanation: 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.

Correct answer: B. (\sqrt{81})

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

Correct answer: A. (\sqrt{81}=9) and the new hypotenuse is (\sqrt{82})

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

Correct answer: A. Because 3² will be added instead of 1²

Explanation: 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.

Correct answer: A. The point will be \(\sqrt{10}\) units away from the origin.

Explanation: 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.

Correct answer: A. Draw a (1) unit perpendicular at the end of (\sqrt{2})

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

Correct answer: B. \(10<\sqrt{120}<11\)

Explanation: \(10^2=100\) 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.

Correct answer: B. √142

Explanation: 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.

Correct answer: A. A perpendicular side of length 1 unit

Explanation: 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.

Correct answer: A. \(5^2<26<6^2\)

Explanation: 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.

Correct answer: A. (√7)²+1²=8

Explanation: 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.

Correct answer: A. (\sqrt{169}=13) and (\sqrt{170}) is irrational

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

Correct answer: A. Take the (\sqrt{n}) hypotenuse length in compass and draw an arc from the origin

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

Correct answer: A. \(\sqrt{56}\), \(7\) और \(8\) के बीच

Explanation: 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.

Correct answer: A. \(1\)

Explanation: 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 \(1\).