Square root spiral

Correct answer: A. (√3)² + 1² = 4

Explanation: The governing concept is the Pythagorean theorem. In the square-root spiral, the existing hypotenuse √3 becomes one perpendicular side of the next right triangle, and a new perpendicular segment of length 1 unit is added. If H is the new hypotenuse, then H² = (√3)² + 1² = 3 + 1 = 4. Taking the positive square root gives H = √4, so option A is the correct calculation. Option C is invalid because lengths cannot be combined in that way to represent a right-triangle hypotenuse. Option B uses a segment of length 3 instead of 1 and gives √12. Option D subtracts squares, whereas the Pythagorean theorem requires addition of the squares of perpendicular sides.

Correct answer: A. Spiral

Explanation: Successive right triangles turn around and make a spiral-like shape. Hence it is called a square root spiral.

Correct answer: C. \(\sqrt{7}\)

Explanation: By Pythagoras, the square of the new hypotenuse is \((\sqrt{6})^2+1^2=6+1=7\). Hence it represents \(\sqrt{7}\). \(\sqrt{6}\) is the previous hypotenuse. Exam tip: square the old length first.

Correct answer: B. (\sqrt{12})

Explanation: Adding a (1) unit perpendicular to (\sqrt{12}) forms (\sqrt{13}). The previous hypotenuse has one less number.

Correct answer: A. पिछले त्रिभुज के कर्ण को एक भुजा तथा 1 इकाई को दूसरी भुजा लेकर

Explanation: In a square root spiral, the previous hypotenuse becomes one side of the next right triangle, and the perpendicular new side is 1 unit. By Pythagoras’ theorem, the new hypotenuse is the square root of the next natural number. Exam tip: identify the added 1-unit side.

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

Explanation: A square root spiral uses right triangles to create successive lengths. If the old hypotenuse is \(\sqrt{n}\) and a new perpendicular of length 1 is drawn at a right angle, the old hypotenuse becomes one leg of the larger right triangle. The new hypotenuse is therefore found from the squares of the two perpendicular sides, not by simply adding lengths.

By Pythagoras, the new length is \(\sqrt{(\sqrt{n})^2+1^2}\). Since \((\sqrt{n})^2=n\) and \(1^2=1\), this becomes \(\sqrt{n+1}\). Thus option A is correct. The other choices either subtract, double, or square the wrong quantity.

Correct answer: A. To make the next square root as hypotenuse

Explanation: A square root spiral is constructed by repeatedly making a right triangle. At each stage, one existing hypotenuse is used as a side, and a new perpendicular side of length 1 unit is drawn. The Pythagorean theorem then gives the length of the new hypotenuse. If the old hypotenuse represents
sqrt{n}
, the new one represents
sqrt{n+1}
.

Thus, adding a 1-unit perpendicular is not meant to erase a line, change a denominator, or turn a triangle into a circle. It creates the next right triangle in the construction. Its hypotenuse has square length
(
sqrt{n}
)^2+1^2=n+1
, so its length is
sqrt{n+1}
. Therefore option A correctly states the purpose: to make the next square root appear as the hypotenuse.

Correct answer: A. The hypotenuse of the previous triangle

Explanation: In a square root spiral, the hypotenuse of one right triangle becomes a side of the next triangle, and a perpendicular side of length 1 is added. Thus, each new hypotenuse represents the next square root. Exam tip: track the previous hypotenuse at every step.

Correct answer: A. √2 and 1

Explanation: Answer: option A, √2 and 1. In a standard square-root spiral, the previous hypotenuse becomes one side of a right triangle, and a new perpendicular segment of length 1 unit is added. To obtain √3, use the preceding hypotenuse √2 together with the unit segment 1. By the Pythagorean theorem, the new hypotenuse h satisfies h² = (√2)² + 1² = 2 + 1 = 3, so h = √3 because lengths are positive. Option B would give √(3+1) = √4, the next stage. Option C gives √8, and option D gives √10. Thus only A produces √3. Memory cue: each step adds one unit square to the square of the previous hypotenuse.

Correct answer: A. By drawing a 1-unit perpendicular side at an end of the previous hypotenuse

Explanation: With previous hypotenuse \(\sqrt{n}\) and a new perpendicular side of 1 unit, Pythagoras gives \(\sqrt{n+1}\). A parallel side does not form the required right triangle. Exam tip: identify the fixed 1-unit leg.

Correct answer: A. By drawing a 1-unit perpendicular segment at the endpoint of the previous hypotenuse

Explanation: The preceding hypotenuse becomes one leg, and a perpendicular unit leg is added at its endpoint. Pythagoras gives next hypotenuse² = previous hypotenuse² + 1. Exam tip: spot the unit perpendicular.

Correct answer: B. 1 unit

Explanation: In a square root spiral, one leg of every new right triangle is fixed at 1 unit, while the other leg is the previous hypotenuse. For example, after \(\sqrt{5}\), the next hypotenuse is \(\sqrt{5+1}=\sqrt{6}\). Exam tip: remember that the fixed leg is always 1 unit.

Correct answer: B. √17

Explanation: The governing rule is that each new hypotenuse is formed by drawing a perpendicular of length 1 unit to the preceding hypotenuse. If the preceding hypotenuse is √n, the Pythagorean theorem gives H² = (√n)² + 1² = n + 1, so the new hypotenuse is √(n + 1). For the target √18, we solve n + 1 = 18 and obtain n = 17. Thus the perpendicular must be drawn on √17, making option B correct. A perpendicular on √16 would produce √17, not √18. One on √18 would produce √19, and one on √19 would produce √20. These distractors represent adjacent stages and therefore test whether the construction rule has been applied in the correct direction.

Correct answer: B. The hypotenuse representing √20

Explanation: In a square-root spiral, successive perpendicular unit segments create right triangles whose hypotenuse lengths are √2, √3, √4, and so on. Continuing the construction until the required stage produces a hypotenuse of length √20. That hypotenuse is transferred with the compass to the number line. Therefore option B is correct; √19 is the preceding stage and 20 units is not the required length.

Correct answer: D. Every hypotenuse is always a whole number.

Explanation: The governing properties of a square-root spiral are that each new triangle is right-angled, a unit perpendicular is added at every stage, and the hypotenuses are √1, √2, √3, √4, and so on. These lengths are not all whole numbers. For example, √1 = 1 and √4 = 2 are whole numbers, but √2, √3, and √5 are irrational and therefore not whole numbers. Consequently, option D is the incorrect statement. Options A, B, and C describe the actual construction: the right angle permits use of the Pythagorean theorem, the added side has unit length, and the resulting hypotenuse sequence represents successive square roots.

Correct answer: A. नए त्रिभुज की एक भुजा के रूप में

Explanation: In a square root spiral, the previous hypotenuse becomes one leg of the next right triangle, while the other leg is 1 unit. By Pythagoras, the new hypotenuse squared is the previous square plus 1. Exam tip: every added triangle is right-angled.

Correct answer: A. 1 unit

Explanation: Each new right triangle in a square root spiral has one leg of length 1 unit and the previous hypotenuse as the other leg. Hence the squares of hypotenuses increase as 2, 3, 4, …. Exam tip: identify the fixed 1-unit leg.

Correct answer: B. √14

Explanation: A square root spiral constructs successive lengths using right triangles. At each step, a new perpendicular side of length 1 is attached to the previously constructed hypotenuse. If the current hypotenuse has length sqrt n, the next one has length sqrt((sqrt n)^2+1^2)=sqrt(n+1) by the Pythagorean theorem. Thus the lengths appear in order as sqrt2,sqrt3,sqrt4,dots.

To construct sqrt15, the immediately preceding step must have produced sqrt14. Adding a unit perpendicular to that length gives sqrt(14+1)=sqrt15. Therefore the length already constructed just before sqrt15 is sqrt14, which is option B. sqrt13 is one step earlier, while sqrt15 is the target itself, not the preceding length.

Correct answer: A. To construct square root lengths

Explanation: A square root spiral geometrically constructs lengths like (\sqrt{2},\sqrt{3},\sqrt{4}). It is also useful on the number line.

Correct answer: A. Pythagoras theorem

Explanation: Pythagoras theorem is used for the hypotenuse in a right triangle. Thus (\sqrt{1^2+1^2}=\sqrt{2}).

Correct answer: A. √n

Explanation: The governing concept is the general recurrence rule of a square-root spiral. A perpendicular segment of length 1 unit is erected on the previous hypotenuse. If the previous hypotenuse is √n, the Pythagorean theorem gives the square of the next hypotenuse as (√n)² + 1² = n + 1. Therefore the next hypotenuse is √(n + 1). Reversing this statement, if the new hypotenuse is √(n + 1), the preceding one must be √n. Hence option A is correct. A previous length √(n + 2) would lead to √(n + 3), while √(2n) would generally lead to √(2n + 1). The expression √(n²) is not the preceding spiral term and does not follow the construction rule.

Correct answer: A. Perpendicular to the previous hypotenuse

Explanation: In a square root spiral, each new 1-unit segment is drawn perpendicular to the previous hypotenuse to form a right triangle. Then the new hypotenuse satisfies h² = n + 1. A parallel segment does not form the required right triangle. Exam tip: check for a 90° angle at every step.

Correct answer: A. नई त्रिभुज की एक भुजा

Explanation: In a square root spiral, the hypotenuse of one triangle becomes one leg of the next right triangle, while the other leg is 1 unit. By Pythagoras, the new hypotenuse becomes \(\sqrt{2}, \sqrt{3}\), and so on. Exam tip: identify the shared side.

Correct answer: A. 1 unit

Explanation: At every step of a square root spiral, a perpendicular side of 1 unit is drawn at an end of the previous hypotenuse. By Pythagoras, (new hypotenuse)² = (previous hypotenuse)² + 1. Exam tip: remember that the fixed added side is always 1 unit.

Correct answer: A. नया त्रिभुज पिछले त्रिभुज के कर्ण पर बनाया जाता है और उसकी एक भुजा 1 इकाई होती है।

Explanation: In a square root spiral, the hypotenuse of the previous triangle becomes one leg of the next right triangle, and the other perpendicular leg is 1 unit. By Pythagoras’ theorem, the new hypotenuse becomes \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. Exam tip: look for the added 1-unit perpendicular side.