Muft Shiksha™ एक 100% Free Education Portal है 🇮🇳, जिसका उद्देश्य Class 9–12 के हर विद्यार्थी तक High-Quality Education को पूरी तरह मुफ्त पहुँचाना है। 🇮🇳 हम मानते हैं कि अच्छी शिक्षा किसी student की आर्थिक स्थिति पर निर्भर नहीं होनी चाहिए। 🇮🇳 हर विद्यार्थी को वही Quality Study Material, MCQs, Quizzes, Exam Preparation, Concept-Based Learning और Bilingual Support मिलना चाहिए, जो आमतौर पर महंगी Coaching या Premium Platforms में मिलता है। Muft Shiksha™ 🇮🇳 इसी सोच के साथ बनाया गया है
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 In a square root spiral, the new hypotenuse is \(\sqrt{n+1}\). If the previous hypotenuse was \(\sqrt{224}\), what will be the new hypotenuse?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: C. \(\sqrt{225}\)
Explanation: The previous hypotenuse is \(\sqrt{224}\), so \(n=224\). In a square root spiral, the next hypotenuse is \(\sqrt{n+1}\); hence it is \(\sqrt{224+1}=\sqrt{225}\). \(\sqrt{224}\) is the previous hypotenuse itself, while \(\sqrt{223}\) belongs to the preceding step. Exam tip: for the next hypotenuse, add 1 to the number inside the square root.
02 When \(\sqrt{225}\) is formed from \(\sqrt{224}\) in a square root spiral, at what value will the new hypotenuse be?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: B. 15
Explanation: In a square root spiral, each new hypotenuse represents \(\sqrt{n}\). Here the new hypotenuse is \(\sqrt{225}\), and since \(225=15^2\), \(\sqrt{225}=15\). Option 14 is incorrect because \(14^2=196\), while \(16^2=256\). Exam tip: first check whether the number under the square root is a perfect square.
03 In a square root spiral, which hypotenuse is formed by drawing a (1) unit perpendicular on \(\sqrt{399}\)?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: B. \(\sqrt{400}\)
Explanation: In each new right triangle of a square root spiral, one side is the previous hypotenuse and the other side is 1 unit. Hence, the new hypotenuse is \(\sqrt{(\sqrt{399})^2+1^2}=\sqrt{399+1}=\sqrt{400}\). \(\sqrt{401}\) would be obtained in the next step, not in this one. Exam tip: at each new step, add 1 to the number under the square root of the hypotenuse.
04 While identifying the interval of \(\sqrt{195}\) in a square root spiral, which conclusion is correct?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: A. \(13<\sqrt{195}<14\)
Explanation: Since \(13^2=169\) and \(14^2=196\), and \(169<195<196\), we get \(13<\sqrt{195}<14\). On the square root spiral, the point for \(\sqrt{195}\) lies between \(\sqrt{169}=13\) and \(\sqrt{196}=14\). Option B is incorrect because \(\sqrt{195}\) is less than 14. Exam tip: Find the nearest smaller and larger perfect squares to determine a square root’s interval quickly.
07 Which statement is correct when comparing \(\sqrt{150}\) and \(\sqrt{169}\) in a square root spiral?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: B. \(\sqrt{150}\) lies between 12 and 13, and \(\sqrt{169}=13\).
Explanation: Since \(12^2=144\) and \(13^2=169\), we have \(144<150<169\). Therefore, \(12<\sqrt{150}<13\). On the other hand, \(169=13^2\), so \(\sqrt{169}=13\). Option A gives the wrong interval for \(\sqrt{150}\). Exam tip: the square root of a number between two consecutive perfect squares lies between their square roots.
08 Why is using (\sqrt{4}) and (1) correct for constructing (\sqrt{5}) in a square root spiral?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: A. Because ((\sqrt{4})^2+1^2=5)
Explanation: The square-root spiral uses a right triangle at every step. To construct sqrt{5}, the previous hypotenuse is sqrt{4} and the newly drawn perpendicular side has length 1. The next hypotenuse is determined by adding the squares of these perpendicular sides, not by adding their lengths directly.
Let the new hypotenuse be h. By Pythagoras, h^2=(sqrt{4})^2+1^2=4+1=5. Because a length is positive, h=sqrt{5}. Therefore option A is correct. The expression sqrt{4}+1 is not sqrt{5}, and (sqrt{4})^2-1^2 gives 3 rather than 5. The construction depends on a sum of squares.
09 In a square root spiral, the hypotenuse formed after \(\sqrt{255}\) will be located at which special value?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: B. \(\sqrt{256}=16\)
Explanation: In a square root spiral, successive hypotenuses are \(\sqrt{1}, \sqrt{2}, \sqrt{3}\), and so on. Hence, the hypotenuse after \(\sqrt{255}\) is \(\sqrt{256}\). Since \(256=16^2\), \(\sqrt{256}=16\), so it lies at an integer value. \(\sqrt{257}\) is the following hypotenuse, while \(\sqrt{254}\) comes earlier. Exam tip: the square root of a perfect square is always an integer.
10 Which statement about the positions of \(\sqrt{24}\) and \(\sqrt{26}\) in a square root spiral is correct?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: A. \(\sqrt{24}\) lies between 4 and 5, whereas \(\sqrt{26}\) lies between 5 and 6
Explanation: Since \(4^2=16\), \(5^2=25\), and \(6^2=36\), \(16<24<25\) gives \(4<\sqrt{24}<5\). Similarly, \(25<26<36\) gives \(5<\sqrt{26}<6\). Hence, the two square roots occur between different consecutive integers on the square root spiral. Option B is a close distractor, but \(\sqrt{26}\) is greater than 5. Exam tip: locate a square root by comparing the number with nearby perfect squares.
11 In a square root spiral, which hypotenuse is formed after \(\sqrt{624}\), and what is its exact value?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: A. \(\sqrt{625}=25\)
Explanation: In a square root spiral, successive hypotenuses are formed as \(\sqrt{2}, \sqrt{3}, \sqrt{4}\), and so on. Therefore, the hypotenuse after \(\sqrt{624}\) is \(\sqrt{625}\). Since \(625=25^2\), \(\sqrt{625}=25\). \(\sqrt{626}\) is the next hypotenuse after it, while \(\sqrt{623}\) is the preceding one. Exam tip: when the number inside a square root is a perfect square, its square root is an integer.
13 In a square root spiral, if (\sqrt{n+1}) is irrational, what conclusion about (n+1) is correct?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: A. (n+1) is not a perfect square
Explanation: The square root of an integer is a whole number only when it is a perfect square. If it is not a perfect square, the square root can be irrational.
14 Which is the correct comparison of \(\sqrt{168}\) and \(\sqrt{170}\) in a square root spiral?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: A. \(\sqrt{168}\) lies between 12 and 13, whereas \(\sqrt{170}\) lies between 13 and 14
Explanation: Here, \(12^2=144\), \(13^2=169\), and \(14^2=196\). Since \(144<168<169\), we get \(12<\sqrt{168}<13\). Similarly, \(169<170<196\) gives \(13<\sqrt{170}<14\). Therefore, option A is correct. Option B is incorrect because \(\sqrt{168}\) is less than 13. Exam tip: To locate a square root, compare the number with the nearest perfect squares.
15 Which option is logical for finding the next hypotenuse from \(\sqrt{12}\) in a square root spiral?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: A. \((\sqrt{12})^2+1^2=13\), so the new hypotenuse is \(\sqrt{13}\)
Explanation: In each new right triangle of a square root spiral, one leg is the previous hypotenuse and the other leg is \(1\). Therefore, by Pythagoras’ theorem, the new hypotenuse is \(\sqrt{(\sqrt{12})^2+1^2}=\sqrt{12+1}=\sqrt{13}\). Option B is incorrect because, in general, \(\sqrt{12}+1\neq\sqrt{13}\). Exam tip: square the previous hypotenuse, add \(1\), and then take the square root.
16 In a square root spiral, which hypotenuse is formed after \(\sqrt{1023}\), and what is its exact value?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: A. \(\sqrt{1024}=32\)
Explanation: In a square root spiral, successive hypotenuses are \(\sqrt{1}, \sqrt{2}, \sqrt{3}\), and so on. Hence, the hypotenuse after \(\sqrt{1023}\) is \(\sqrt{1024}\). Since \(1024=32^2\), \(\sqrt{1024}=32\). \(\sqrt{1025}\) is the following hypotenuse, not the immediate next one. Exam tip: remember nearby perfect squares; \(32^2=1024\).
19 Which statement is most incorrect from the construction point of view in a square-root spiral?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: D. Finding the next hypotenuse by directly adding 1 to the previous hypotenuse
Explanation: The governing concept is the Pythagorean construction of successive hypotenuses. At every stage, a unit segment is drawn perpendicular to the previous hypotenuse, which becomes a side of the new right triangle. Consequently, if the previous hypotenuse is √n, the next one is √(n+1), obtained from √[(√n)²+1²], not by ordinary addition. Thus option D is the most incorrect statement. Options A, B and C describe essential construction steps: a right angle is required, the added perpendicular side has unit length, and the preceding hypotenuse is reused. Directly writing √n+1 would generally give a different value from √(n+1).
22 In a square root spiral, if the previous hypotenuse is \(\sqrt{483}\), at what exact value will the new hypotenuse be after adding a (1) unit perpendicular?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: A. \(\sqrt{484}=22\)
Explanation: In each new right triangle of a square root spiral, one perpendicular side has length 1. By the Pythagorean theorem, the square of the new hypotenuse is \(483+1=484\). Hence, the new hypotenuse is \(\sqrt{484}=22\). \(\sqrt{485}\) is not correct because adding a unit side increases the square of the hypotenuse by 1, not the hypotenuse itself. Exam tip: add 1 to the number under the previous radical, then take the square root.
23 If the new hypotenuse in a square root spiral is equal to (31), which hypotenuse came immediately before it?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: A. \(\sqrt{960}\)
Explanation: In a square root spiral, successive hypotenuses are of the form \(\sqrt{n}\), with the number inside the square root increasing by 1 at each step. The given new hypotenuse is \(31=\sqrt{961}\). Therefore, the immediately preceding hypotenuse is \(\sqrt{960}\). \(\sqrt{961}\) is the given current hypotenuse, not the previous one. Exam tip: Express an integral hypotenuse as its square root form first, then subtract 1 from the radicand for the preceding step.
24 Before placing \(\sqrt{1155}\) on the number line, which interval will be correctly identified?
0 reads0 helpful★ – (0)
Answer and explanation
Correct answer: B. \(33<\sqrt{1155}<34\)
Explanation: We have \(33^2=1089\) and \(34^2=1156\). Since \(1089<1155<1156\), taking square roots gives \(33<\sqrt{1155}<34\). Although it is very close to \(34\), \(\sqrt{1155}\) is less than \(34\) because \(1155<34^2\). Exam tip: To locate a square root, find the nearest perfect squares on either side of the given number.
☆No ratings yetWrite a review / Rate this question
Was this question useful?
👍 0 Helpful ·👎 0 Not helpful
Difficulty
Easy0%
Medium0%
Hard0%
Was the explanation clear?
Yes 0%·No 0%
0 responses
Student Reviews
No published reviews yet.
Analytics choices
Google Analytics helps us understand site usage. Google may send limited cookie-free signals before your choice. The Live Visitors widget operates independently of this analytics choice; see the privacy policy for its provider and fallback details. Essential site features work without analytics cookies. You can change your choice later in Privacy choices. Privacy policy