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In this Class 9 Mathematics topic from Number Systems, students explore the square root spiral, a geometric construction that represents √2, √3, √4 and successive square-root lengths. They learn how right triangles and the Pythagorean theorem generate each new radius, connect these constructions with irrational numbers, and locate their values on the number line. The topic strengthens understanding of square roots, geometric representation, measurement, and the relationship between numerical patterns and visual reasoning.
Practice questions
01 To construct √224 in a square root spiral, what is the correct previous hypotenuse and in which interval will the formed hypotenuse lie?
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Answer and explanation
Correct answer: B. √223, between 14 and 15
Explanation: Each step of the square root spiral adds 1 to the squared hypotenuse, so a target √224 must be constructed from the previous hypotenuse √223 with a new perpendicular of 1 unit: (√223)² + 1² = 224. To locate the result, compare 224 with consecutive perfect squares. Since 14² = 196 and 15² = 225, we have 196 < 224 < 225. Taking positive square roots gives 14 < √224 < 15. Therefore option B is correct. Option A has the wrong predecessor, option C reverses the correct interval, and option D would apply to √225, not √224.
02 After obtaining \(\sqrt{18}\) using the square root spiral, Riya marks it at 4.5 on the number line. Which comment about her marking is correct?
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Answer and explanation
Correct answer: A. Riya’s marking is incorrect; \(\sqrt{18}\) lies between 4 and 5 and is closer to 4.
Explanation: Since \(4^2=16\) and \(5^2=25\), \(\sqrt{18}\) must lie between 4 and 5. Its value is about 4.24, while \(4.5^2=20.25\). Exam tip: compare the nearest perfect squares first.
04 A student says that a base of length 10 units must first be drawn to represent \(\sqrt{10}\) on the square root spiral. Which option correctly explains the student's error?
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Answer and explanation
Correct answer: B. Adding a perpendicular side of 1 unit to the side of length \(\sqrt{9}\) gives a hypotenuse of length \(\sqrt{10}\).
Explanation: Each new right triangle in the square root spiral has a perpendicular side of 1 unit. \(\sqrt{9^2+1^2}=\sqrt{10}\), so a 10-unit base is not needed. Exam tip: apply Pythagoras' theorem.
05 Why are (\sqrt{4}) and (1) correct sides for constructing (\sqrt{5}) in a square root spiral?
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Answer and explanation
Correct answer: A. Because ((\sqrt{4})^2+1^2=5)
Explanation: The direct answer is A. In a square-root spiral, the old hypotenuse becomes one side of a new right triangle, and a perpendicular side of length 1 is added. Here the old length is sqrt{4}. By Pythagoras, the square of the new hypotenuse is (sqrt{4})^2+1^2=4+1=5, so the new hypotenuse is sqrt{5}. Option A is correct because it uses squares of the perpendicular sides. Option B is wrong: lengths are not added directly, since sqrt{4}+1=3, not sqrt{5}. Option C is wrong because 5 is not equal to sqrt{5}. Option D is wrong because the Pythagorean theorem uses addition for a right triangle, not subtraction. Remember: square, add, then take the square root.
06 Before placing \(\sqrt{440}\) on the number line using a square root spiral, which interval is correct?
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Answer and explanation
Correct answer: B. \(20<\sqrt{440}<21\)
Explanation: \(20^2=400\) and \(21^2=441\). Since \(400<440<441\), taking square roots gives \(20<\sqrt{440}<21\). Although it is close to \(21\), \(\sqrt{440}\) is less than \(21\) because \(440<441\). Exam tip: To locate a square root, compare the number with consecutive perfect squares.
07 If the (k)-th hypotenuse in a square root spiral is considered as (\sqrt{k}), which hypotenuse is (\sqrt{25}), and what is its value?
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Answer and explanation
Correct answer: B. (25)-th, (5)
Explanation: The notation in this question directly defines the kth hypotenuse as \\(\\sqrt{k}\\). Therefore, the number under the radical identifies the position of the hypotenuse, while evaluating the radical gives its numerical length. For \\(\\sqrt{25}\\), the index is 25 and the length is 5. Hence option B contains both required parts correctly.
The calculation is \\(\\sqrt{25}=5\\), because 25 is the square of 5. The phrase “25th hypotenuse” refers to the position in the spiral, not to a length of 25. Thus option D confuses the index with the value. The 24th and 26th positions would represent \\(\\sqrt{24}\\) and \\(\\sqrt{26}\\), respectively, so options A and C do not describe the stated segment.
08 A student says that in a square root spiral, the length of the next hypotenuse after \(\sqrt{14}\) will be \(\sqrt{14}+1\). Which reasoning correctly shows the error in this statement?
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Answer and explanation
Correct answer: A. The square of the new hypotenuse is \(14+1=15\), so its length is \(\sqrt{15}\).
Explanation: Apply Pythagoras to the new hypotenuse: \(h^2=14+1=15\), so \(h=\sqrt{15}\). Directly adding 1 to \(\sqrt{14}\) is invalid. Exam tip: add squares, not lengths.
12 In a square root spiral, if the hypotenuse (\sqrt{n+1}) formed after (\sqrt{n}) is irrational, what can be definitely true?
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Answer and explanation
Correct answer: A. (n+1) is not a perfect square
Explanation: The square root of a positive integer is a whole number only when it is a perfect square. If it is not a perfect square, the square root is irrational.
14 Which feature of the construction of a square root spiral is correct?
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Answer and explanation
Correct answer: A. In each new right triangle, one leg is the hypotenuse of the previous triangle and the other leg is 1 unit.
Explanation: Each new right triangle uses the previous hypotenuse as one leg and a unit length as the other, producing \(\sqrt{2}, \sqrt{3}\), and so on. Exam tip: identify the newly added 1-unit leg first.
16 While constructing a square root spiral, Reema used a new side of
√2
units to make the triangle after the point representing
√7
. What is her mistake?
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Answer and explanation
Correct answer: A. The new perpendicular side should be 1 unit, not
√2
units.
Explanation: In a square root spiral, the newly added perpendicular side is always 1 unit. After
√7
, the hypotenuse is
√(7+1)=√8
. Using
√2
would produce
√9
instead. Exam tip: apply the fixed 1-unit side rule at every step.
19 In a square root spiral, point P corresponding to \(\sqrt{18}\) is marked. Riya says that P will lie between 4 and 5 units from the centre O. What is the correct evaluation of Riya's statement?
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Answer and explanation
Correct answer: A. Riya is correct because \(OP=\sqrt{18}\) and \(4^2<18<5^2\).
Explanation: In a square root spiral, the distance OP is \(\sqrt{18}\). Since \(4^2=16\) and \(5^2=25\), \(4<\sqrt{18}<5\). Option B is wrong because \(3^2=9\). Exam tip: compare with nearby perfect squares.
20 In a standard square root spiral, which of the following lengths is not obtained as a radial segment?
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Answer and explanation
Correct answer: D. \(\frac{3}{2}\)
Explanation: Every radial segment in a square root spiral has length \(\sqrt{n}\), where \(n\) is a positive integer. \(\sqrt{2}\), \(\sqrt{12}\), and \(\sqrt{81}\) fit this form, but \(\frac{3}{2}\) does not. Exam tip: test whether the length can be written as \(\sqrt{n}\).
23 Which conclusion is correct for locating the point at a distance \(\sqrt{50}\) on the number line using a square root spiral?
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Answer and explanation
Correct answer: A. \(\sqrt{50}\) lies between 7 and 8 because \(49<50<64\)
Explanation: Since \(7^2=49\) and \(8^2=64\), \(49<50<64\) gives \(7<\sqrt{50}<8\). Hence A is correct; \(\sqrt{50}\) is not exactly 7. In exams, compare the nearest perfect squares.
24 If the new hypotenuse in a square root spiral is equal to (14), which hypotenuse came immediately before it?
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Answer and explanation
Correct answer: A. \(\sqrt{195}\)
Explanation: In a square root spiral, the successive hypotenuses are \(\sqrt{1}, \sqrt{2}, \sqrt{3}, \ldots\). The given new hypotenuse is \(14\), and \(14=\sqrt{196}\). Therefore, the hypotenuse immediately before it is \(\sqrt{195}\). \(\sqrt{196}\) is the current hypotenuse, not the preceding one. Exam tip: Express the whole number as \(\sqrt{n}\), then subtract 1 from the radicand to find the previous hypotenuse.
25 A student has constructed a square root spiral up to \(\sqrt{12}\). To obtain \(\sqrt{13}\), she draws a perpendicular of length 1 unit at the end of the last radius. Which conclusion is correct?
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Answer and explanation
Correct answer: A. The new hypotenuse will have length \(\sqrt{13}\), because \((\sqrt{12})^2+1^2=13\).
Explanation: In each new right triangle, the previous hypotenuse is one leg and 1 is the other. Thus, new hypotenuse² = 12 + 1 = 13, so its length is \(\sqrt{13}\). Adding \(\sqrt{12}+1\) is incorrect. Exam tip: apply Pythagoras’ theorem.
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