Square root spiral

Correct answer: B. इसका वर्ग पिछली कर्ण के वर्ग से 1 अधिक होता है।

Explanation: By Pythagoras’ theorem, \(h_{new}^2=h_{old}^2+1^2\). Thus, the square of the new hypotenuse increases by 1, giving \(\sqrt2,\sqrt3,\sqrt4\) in order. Exam tip: write the squares first to avoid adding lengths directly.

Correct answer: A. A line segment of length 1 is drawn perpendicular to the previous hypotenuse

Explanation: In a square root spiral, the previous hypotenuse becomes one leg of the next right triangle, and a unit segment is drawn perpendicular to it. Hence the new hypotenuse represents the next square root. Exam tip: look for the perpendicular unit side.

Correct answer: B. ((\sqrt{3})^2+1^2=4), so the hypotenuse is (\sqrt{4})

Explanation: In a square root spiral, the sum of squares is taken by Pythagoras theorem. Directly adding lengths is wrong.

Correct answer: A. (\sqrt{225}), (15)

Explanation: In the square root spiral, each new right triangle has the previous hypotenuse as one side and a new perpendicular side of length 1. If the current hypotenuse is \(\sqrt{224}\), then the next one has square equal to \((\sqrt{224})^2+1^2=224+1=225\). Therefore the next hypotenuse is \(\sqrt{225}\). This follows directly from the Pythagorean theorem and the fixed unit segment used in the spiral.

Since 225 is a perfect square, \(\sqrt{225}=15\), because \(15^2=225\). Thus option A gives both the correct next hypotenuse and its exact value. Option D has the right radical but the wrong value, since 14 squared is 196, not 225. Options B and C do not follow the successive construction rule. The supplied answer is therefore correct.

Correct answer: B. Both are irrational numbers

Explanation: Neither \(2\) nor \(8\) is a perfect square. The square root of a positive integer is rational only when the integer is a perfect square. Hence \(\sqrt{2}\) and \(\sqrt{8}=2\sqrt{2}\) are both irrational. Option C is incorrect because \(\sqrt{2}\) is not a whole number. Exam tip: First check whether the number under the square root is a perfect square.

Correct answer: A. Pythagoras' theorem

Explanation: By Pythagoras' theorem, if the square of the previous hypotenuse is \(n\) and the new perpendicular side is 1, the new hypotenuse has square \(n+1\). Hence it represents \(\sqrt{n+1}\). Thales' theorem is not used here. Exam tip: identify the right angle first.

Correct answer: B. \(\sqrt{4}\) and \(1\)

Explanation: To obtain \(\sqrt{5}\) in a square root spiral, the previous hypotenuse \(\sqrt{4}\) is taken perpendicular to a unit side. By the Pythagorean theorem, the square of the new hypotenuse is \((\sqrt{4})^2+1^2=4+1=5\), so the hypotenuse is \(\sqrt{5}\). In option A, the hypotenuse would be \(\sqrt{3+4}=\sqrt{7}\), not \(\sqrt{5}\). Exam tip: At each new step, use the previous square root and \(1\) as perpendicular sides.

Correct answer: A. The other perpendicular side should be 1 unit.

Explanation: In a square root spiral, the previous hypotenuse \(\sqrt{7}\) is retained as one leg and the other leg is always 1 unit. Hence the new hypotenuse is \(\sqrt{7+1}=\sqrt{8}\). Taking \(\sqrt{7}\) as the other leg gives \(\sqrt{14}\), not \(\sqrt{8}\). Exam tip: always check the unit leg.

Correct answer: D. \(90^\circ\)

Explanation: Each new triangle in a square root spiral is right-angled, so the new 1-unit side is drawn perpendicular to the previous hypotenuse. If the old hypotenuse is \(\sqrt n\), the new one becomes \(\sqrt{n+1}\). Exam tip: perpendicular always means \(90^\circ\).

Correct answer: A. They allow (√n)² + 1² = n + 1 to apply

Explanation: The governing idea of a square root spiral is the repeated use of a right triangle. If the existing hypotenuse has length √n, a new perpendicular segment of length 1 is drawn at a right angle to it. By the Pythagorean theorem, the new hypotenuse has squared length (√n)² + 1² = n + 1, so its length is √(n + 1). Repeating this construction produces √2, √3, √4 and so on. Thus option A correctly identifies both essential conditions. Option B is false because the hypotenuse generally increases, option C is false because most roots remain irrational, and option D is false because a right triangle is not equilateral.

Correct answer: A. उसकी मूलबिंदु से दूरी \(\sqrt{n}\) होती है।

Explanation: In a square root spiral, each new right triangle adds a side of length 1. By Pythagoras, the new squared distance is \((n-1)+1=n\), so the radius is \(\sqrt{n}\). Exam tip: identify numbers by their distance from the origin, not by the angle.

Correct answer: D. Finding the next hypotenuse by directly adding (1) to the previous hypotenuse

Explanation: Direct addition is not used in a square root spiral. The new hypotenuse is obtained from a right triangle and Pythagoras theorem.

Correct answer: B. (\sqrt{223})

Explanation: Adding a (1) unit perpendicular to (\sqrt{223}) forms (\sqrt{224}). The number increases by (1) in the next hypotenuse.

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

Explanation: The hypotenuse length of the square root to be marked is taken in the compass. Drawing an arc from the origin is the correct way.

Correct answer: A. One leg \(\sqrt{9}\), the other leg \(1\), and hypotenuse \(\sqrt{10}\)

Explanation: In the spiral, each new hypotenuse is formed using the previous hypotenuse and a perpendicular unit side. \((\sqrt{9})^2+1^2=10\), so it is \(\sqrt{10}\). \(\sqrt{10}\) instead gives \(\sqrt{11}\). Tip: track the hypotenuse.

Correct answer: A. \(\sqrt{2}\)

Explanation: The first right triangle has two perpendicular sides of 1 unit each. Hence its hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). \(\sqrt{3}\) occurs in the next triangle. Exam tip: remember that the spiral begins with \(\sqrt{2}\).

Correct answer: A. \(\sqrt{24}\) is between \(4\) and \(5\), and \(\sqrt{25}=5\)

Explanation: Since \(4^2=16\), \(5^2=25\), and \(16<24<25\), we get \(4<\sqrt{24}<5\). Also, \(25=5^2\) is a perfect square, so \(\sqrt{25}=5\). Option B is wrong because \(\sqrt{24}\) is not equal to \(5\). Exam tip: compare a number with nearby perfect squares to locate its square root.

Correct answer: A. The hypotenuse of a triangle with perpendicular sides \(\sqrt{12}\) and 1

Explanation: In a square root spiral, drawing a perpendicular unit segment at \(\sqrt{12}\) gives \(\sqrt{(\sqrt{12})^2+1^2}=\sqrt{13}\). A number need not be a perfect square. In exams, apply Pythagoras’ theorem.

Correct answer: A. A unit-length side is drawn perpendicular to the previous hypotenuse at its endpoint

Explanation: The previous hypotenuse becomes one leg, and a perpendicular unit leg is added at its endpoint. By Pythagoras, if its square is n, the new hypotenuse has square n+1. A parallel line will not form the required right triangle. Exam tip: remember “perpendicular + 1 unit.”

Correct answer: B. √109 and 1 unit

Explanation: In the standard square root spiral, each new right triangle has a perpendicular of length 1 unit. If the previous hypotenuse is √k, the Pythagorean theorem gives the next hypotenuse as √(k + 1), because (√k)² + 1² = k + 1. To obtain √110, we therefore need k + 1 = 110, so k = 109. The preceding hypotenuse must be √109, followed by a 1-unit perpendicular at a right angle. Option B is correct. Option A would produce √109 rather than √110, while options C and D use the wrong starting value or move in the wrong numerical direction.

Correct answer: A. पिछले कर्ण के एक सिरे पर 1 इकाई का लंब खींचा जाता है।

Explanation: In a square root spiral, a 1-unit perpendicular is drawn at an endpoint of the previous hypotenuse to form the next right triangle. By Pythagoras, the new hypotenuse becomes \(\sqrt{n+1}\). A parallel line would not create a right angle. Exam tip: each new outer leg is 1 unit.

Correct answer: A. \(\sqrt{2}\rightarrow\sqrt{3}\rightarrow\sqrt{4}\rightarrow\sqrt{5}\)

Explanation: In a square root spiral, hypotenuses are formed successively. In the usual construction, intermediate steps are not skipped.

Correct answer: A. It is a chain of right triangles where the previous hypotenuse and (1) unit perpendicular form the next square root

Explanation: The direct answer is A. A square-root spiral is a chain of right triangles. Begin with a suitable right triangle, and at each stage draw a new perpendicular side of length 1 unit to the previous hypotenuse. By Pythagoras, if the old hypotenuse is sqrt{n}, the new one has length sqrt{n+1}, because (sqrt{n})^2+1^2=n+1. Thus option A gives the essential construction. Option B is wrong because the spiral represents successive square roots, including non-perfect-square roots, not merely a list of perfect squares. Option C is wrong because square roots are not added directly; the Pythagorean relation is used. Option D is wrong because circles are not the defining construction. Memory cue: one new unit perpendicular plus the old hypotenuse gives the next square root.

Correct answer: A. Draw a perpendicular of length 1 at the endpoint of \(\sqrt{7}\), and join its new endpoint to the initial point

Explanation: In a square root spiral, a unit perpendicular at the endpoint of \(\sqrt{7}\) gives a new hypotenuse of \(\sqrt{7+1}=\sqrt{8}\). A number need not be a perfect square. Exam tip: each new hypotenuse represents \(\sqrt{n+1}\).

Correct answer: B. (\sqrt{324}), (18)

Explanation: The new hypotenuse is (\sqrt{323+1}=\sqrt{324}), and (\sqrt{324}=18). When a perfect square appears, write the exact whole number.