If the new hypotenuse obtained at a step is (\sqrt{42}), which was the immediately previous hypotenuse?
Correct answer: B
In the previous step, the number under the root is (1) less. Therefore, (\sqrt{41}) comes before (\sqrt{42}).
Mathematics
Square root spiral
वर्गमूल सर्पिल
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.
TOPIC PRACTICE
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Expert · Level 3
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- (\sqrt{40})
- (\sqrt{41})
- (\sqrt{42})
- (\sqrt{43})
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- (46)
- (47)
- (48)
- (49)
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- (\sqrt{22},\sqrt{23},\sqrt{24})
- (\sqrt{22},\sqrt{24},\sqrt{26})
- (\sqrt{21},\sqrt{23},\sqrt{25})
- (\sqrt{20},\sqrt{21},\sqrt{23})
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- The newly added side has length 1 and is perpendicular to the previous hypotenuse.
- The newly added side has length 1 and is parallel to the previous hypotenuse.
- The two perpendicular sides of every new triangle are equal in length.
- The hypotenuse of every new triangle is always 1 unit long.
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- (\sqrt{29})
- (\sqrt{30})
- (\sqrt{31})
- (\sqrt{32})
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- The difference of lengths is not (1), the difference of their squares is (1)
- Both lengths are equal
- The difference of their squares is (2)
- (\sqrt{11}) is not formed in the spiral
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- (6\sqrt{2})
- (8\sqrt{2})
- (12\sqrt{2})
- (36\sqrt{2})
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- (\sqrt{45})
- (\sqrt{49})
- (\sqrt{52})
- (\sqrt{60})
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- (\sqrt{79})
- (\sqrt{80})
- (\sqrt{81})
- (\sqrt{82})
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- \(\sqrt{a-1}\)
- \(\sqrt{a}\)
- \(\sqrt{a+1}\)
- \(a+1\)
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- (\sqrt{97}) and (\sqrt{100})
- (\sqrt{98}) and (\sqrt{100})
- (\sqrt{99}) and (\sqrt{100})
- (\sqrt{98}) and (\sqrt{101})
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- 10
- 11
- 12
- 121
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- \(\sqrt{20}\)
- \(\sqrt{21}\)
- \(\sqrt{22}\)
- \(\sqrt{23}\)
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- \(2\sqrt{27}\)
- \(3\sqrt{6}\)
- \(6\sqrt{3}\)
- \(9\sqrt{6}\)
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- Making every angle (60^\circ)
- Drawing a (1) unit perpendicular to the previous hypotenuse
- Drawing a parallel line each time
- Drawing only circles from the center
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- Yes, every hypotenuse will be at (45^\circ)
- No, the direction changes because each new perpendicular is drawn on the previous hypotenuse
- Yes, because all sides are equal
- No, because (\sqrt{2}) is not formed
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- Yes, because (1) is added
- No, because a right angle is necessary for Pythagoras theorem
- Yes, because every slant line is perpendicular
- No, because (\sqrt{16}) is not formed in the spiral
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- 2
- 3
- 4
- 5
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- \(3\sqrt{10}\)
- \(9\sqrt{10}\)
- \(10\sqrt{9}\)
- \(5\sqrt{6}\)
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- (10)
- (\sqrt{100})
- (\sqrt{101})
- (101)
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- The distance obtained from the spiral can be marked on the number line using a compass
- The spiral shows only negative numbers
- Square roots formed in the spiral cannot be placed on the number line
- The spiral replaces the number line
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- \(\sqrt{61}\)
- \(\sqrt{62}\)
- \(\sqrt{63}\)
- \(\sqrt{64}\)
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- Length of every new perpendicular segment is (1)
- Length of the initial segment is (1)
- Length of every new perpendicular segment is (2)
- The first right triangle is made with (1) and (1)
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- (2\sqrt{30})
- (4\sqrt{15})
- (6\sqrt{20})
- (10\sqrt{12})
Expert · Level 3
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- \(2\sqrt{30}\)
- \(4\sqrt{15}\)
- \(6\sqrt{20}\)
- \(10\sqrt{12}\)
If the spiral starts from (\sqrt{1}) and is constructed up to (\sqrt{48}), how many right triangles are formed?
Correct answer: B
(\sqrt{2}) is given by the first triangle, so up to (\sqrt{48}), the number of triangles is (48-1=47). Keep the initial (\sqrt{1}) separate in counting.
Which option gives the correct order of three consecutive hypotenuses?
Correct answer: A
In consecutive hypotenuses, the numbers under the roots increase by (1). Therefore, (\sqrt{22},\sqrt{23},\sqrt{24}) is the correct order.
Which statement is correct for constructing each new right triangle in a square root spiral?
Correct answer: A
In a square root spiral, a unit side is drawn perpendicular to the preceding hypotenuse. By Pythagoras, \((\sqrt{n-1})^2+1^2=n\), so the new hypotenuse is \(\sqrt{n}\). Exam tip: identify both the unit side and the right angle.
To construct (\sqrt{31}) in a square root spiral, on which hypotenuse is a perpendicular of (1) unit drawn?
Correct answer: B
((\sqrt{30})^2+1^2=31), so the perpendicular is drawn on (\sqrt{30}). For the next square root, identify the previous inner value.
If a student says that the difference between the lengths (\sqrt{10}) and (\sqrt{11}) is (1), what is the correct correction?
Correct answer: A
((\sqrt{11})^2-(\sqrt{10})^2=1), but (\sqrt{11}-\sqrt{10}\neq1). In exams, understand the difference between length and square.
When (\sqrt{72}) is obtained in the spiral, what is its simplest form?
Correct answer: A
The direct answer is A: \\(6\\sqrt{2}\\). Simplify the radical by looking for a perfect-square factor: \\(72=36\\times2\\). Therefore, \\(\\sqrt{72}=\\sqrt{36\\times2}=\\sqrt{36}\\sqrt{2}=6\\sqrt{2}\\). The square-root spiral does not change this algebraic simplification. A is correct. B, \\(8\\sqrt{2}\\), is wrong because its square is \\(128\\), not 72. C, \\(12\\sqrt{2}\\), is wrong because its square is \\(288\\). D, \\(36\\sqrt{2}\\), is wrong because its square is \\(2592\\). The error in the wrong choices is taking an incorrect multiplier outside the root. Memory cue: remove the largest perfect-square factor; \\(\\sqrt{36}=6\\).
Which hypotenuse will be formed in the spiral and have a natural number value?
Correct answer: B
The direct answer is option B: \(\sqrt{49}=7\). A natural number is a positive whole number such as 1, 2, 3, and so on. In the square-root spiral, the successive hypotenuse lengths are square roots of positive integers. A square root is a natural number when its radicand is a perfect square. Check each option: \(\sqrt{45}=3\sqrt{5}\), not a natural number, because 45 is not a perfect square. \(\sqrt{49}=7\), and 7 is a natural number, so option B works. \(\sqrt{52}=2\sqrt{13}\), not a natural number. \(\sqrt{60}=2\sqrt{15}\), also not a natural number. The fact that these lengths can occur in the spiral does not make every one of them a whole number. Exam cue: test whether the number under the radical is a perfect square; 49 is \(7^2\).
If the spiral is constructed up to (\sqrt{80}), which hypotenuse will be formed immediately after it?
Correct answer: C
In the next step, the inner number increases by (1), so (\sqrt{81}) is formed. The spiral does not skip numbers in order.
If \(OP=\sqrt{a}\) and \(PQ=1\) is perpendicular, what is the general form of the next hypotenuse (OQ)?
Correct answer: C
In the standard square-root spiral construction, \(PQ\) is perpendicular to \(OP\). By Pythagoras’ theorem, \(OQ^2=OP^2+PQ^2=(\sqrt{a})^2+1^2=a+1\). Hence, \(OQ=\sqrt{a+1}\). The expression \(a+1\) is the value of \(OQ^2\), not the length of the hypotenuse itself. Exam tip: add the squares of the perpendicular sides first, then take the square root.
In a square root spiral, which hypotenuses come immediately before and after (\sqrt{99})?
Correct answer: B
In order, the inner number decreases by (1) before and increases by (1) after. So (\sqrt{98}) comes before (\sqrt{99}), and (\sqrt{100}) comes after it.
If \(\sqrt{121}\) is formed in the spiral, how many units is it in actual length?
Correct answer: B
\(121\) is a perfect square because \(11 \times 11 = 121\). Therefore, \(\sqrt{121}=11\), so its actual length in the spiral is 11 units. \(121\) is the number under the square root, not its square root. Exam tip: Memorising squares from 1 to 15 helps solve such questions quickly.
If after (7) new steps the hypotenuse becomes \(\sqrt{29}\), what was the starting hypotenuse?
Correct answer: C
In a square root spiral, each new step increases the number under the square root of the hypotenuse by 1. Since 7 new steps lead to \(\sqrt{29}\), the starting number was \(29-7=22\). Hence, the starting hypotenuse was \(\sqrt{22}\). If it had been \(\sqrt{21}\), seven steps would give \(\sqrt{28}\), not \(\sqrt{29}\). Exam tip: Add or subtract the number of steps from the radicand, not from the square root itself.
Which option gives the correct simplest form of \(\sqrt{54}\)?
Correct answer: B
Since \(54=9\times6\), and \(9\) is the greatest perfect-square factor of 54, \(\sqrt{54}=\sqrt{9\times6}=\sqrt9\times\sqrt6=3\sqrt6\). The expression \(2\sqrt{27}\) is neither in simplest form nor equal to \(\sqrt{54}\). Exam tip: first identify the greatest perfect-square factor inside the radical.
Which geometric idea is most necessary instead of an angle protractor in constructing a square root spiral?
Correct answer: B
The main construction depends on drawing a (1) unit perpendicular to the previous hypotenuse. This creates the next right triangle.
If the hypotenuse (\sqrt{2}) appears in the (45^\circ) direction, will every next hypotenuse also be at (45^\circ)?
Correct answer: B
At each new step, the perpendicular is drawn on the previous hypotenuse, so the direction gradually changes. The spiral is formed by this turning.
If (OP=\sqrt{15}), (PQ=1), and (PQ) is slanting instead of perpendicular, is (OQ=\sqrt{16}) certain?
Correct answer: B
(\sqrt{16}) is certain only when the new segment is perpendicular to the previous hypotenuse. Without a right angle, Pythagoras theorem cannot be applied.
How many new steps are needed to go from \(\sqrt{3}\) to \(\sqrt{7}\) in the spiral?
Correct answer: C
In a square root spiral, each new step represents the next number: \(\sqrt{1}, \sqrt{2}, \sqrt{3}, \ldots\). From \(\sqrt{3}\) to \(\sqrt{7}\), the radicand changes from 3 to 7, so the number of new steps is \(7-3=4\). The four steps correspond to \(\sqrt{4}, \sqrt{5}, \sqrt{6}\), and \(\sqrt{7}\). Exam tip: subtract the starting radicand from the ending radicand in such questions.
If \(\sqrt{90}\) is obtained in the spiral, what is its simplest form?
Correct answer: A
Since \(90=9\times10\) and \(9\) is a perfect square, \(\sqrt{90}=\sqrt{9\times10}=\sqrt9\sqrt{10}=3\sqrt{10}\). \(9\sqrt{10}\) is incorrect because \(\sqrt9=3\), not 9. Exam tip: identify the greatest perfect-square factor when simplifying a surd.
If one hypotenuse is (\sqrt{100}) and the next step is constructed, what will be the actual value of the new hypotenuse?
Correct answer: C
The next hypotenuse is (\sqrt{101}), which is not a simple integer. Even after (\sqrt{100}=10), the sequence continues with (\sqrt{101}).
Which statement correctly describes the relation between a square root spiral and the number line?
Correct answer: A
The spiral gives an exact distance that can be transferred to the number line with a compass. It is useful for showing irrational numbers.
If hypotenuses are constructed up to \(\sqrt{65}\), which hypotenuse was (3) steps before the final one?
Correct answer: B
In a square root spiral, the number under the radical increases by 1 at each successive step. Going 3 steps back from \(\sqrt{65}\) gives \(65-3=62\). Hence, the required hypotenuse is \(\sqrt{62}\). \(\sqrt{63}\) is only 2 steps before the final hypotenuse. Exam tip: subtract the number of backward steps from the radicand.
Which option gives an incorrect fixed length for constructing the spiral?
Correct answer: C
A square-root spiral is built by repeatedly drawing a new segment perpendicular to the previous segment. In the standard construction, each newly drawn perpendicular segment has a fixed length of 1 unit. The initial segment also has length 1 unit, and these two perpendicular unit segments form the first right triangle. Its hypotenuse is therefore (1^2+1^2)=\sqrt{2}, which begins the sequence of square roots.
Thus, the statements saying that the initial segment is 1 unit, every new perpendicular segment is 1 unit, and the first triangle uses two segments of length 1 are consistent with the standard construction. Choosing a new perpendicular segment of length 2 changes the Pythagorean calculation and produces a different construction, not the usual square-root spiral. Therefore option C is the incorrect fixed length. The supplied answer and explanation are accurate.
If (OP=\sqrt{120}), what is the simplest form of (OP)?
Correct answer: B
(\sqrt{120}=\sqrt{4\times30}=2\sqrt{30}) is simplest, while (4\sqrt{15}) squares to (240). Check each option carefully.
If \(OP=\sqrt{120}\), which is the correct simplest form of (OP)?
Correct answer: A
Since \(120=4\times30\), and \(4\) is a perfect square, \(\sqrt{120}=\sqrt{4\times30}=2\sqrt{30}\). As 30 has no perfect-square factor greater than 1, this is the simplest form. The close distractor \(4\sqrt{15}\) is incorrect because its square is \(16\times15=240\), not 120. Exam tip: identify the greatest perfect-square factor inside the radical before simplifying.
QUIZ COMPLETE