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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.
TOPIC PRACTICE
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Hard · Level 6View options
\(\sqrt{1598}\)
\(\sqrt{1600}\)
\(\sqrt{3198}\)
\(\sqrt{1601}\)
Hard · Level 6View options
\(35<\sqrt{1368}<36\)
\(36<\sqrt{1368}<37\)
\(37<\sqrt{1368}<38\)
\(\sqrt{1368}=37\)
Hard · Level 6View options
Take the (\sqrt{3}) hypotenuse length in a compass and draw an arc from the origin
Directly mark (3) units
Draw any arc from any point
Mark double the hypotenuse
Hard · Level 6View options
(\sqrt{838})
(\sqrt{839})
(\sqrt{840})
(\sqrt{841})
Hard · Level 6View options
\(\sqrt{960}\) lies between 30 and 31, and \(\sqrt{1024}=32\).
\(\sqrt{960}>31\), and \(\sqrt{1024}=32\).
\(\sqrt{960}=31\), and \(\sqrt{1024}\) is irrational.
\(\sqrt{960}=32\), and \(\sqrt{1024}>32\).
Hard · Level 6View options
Because ((\sqrt{9})^2+1^2=10)
Because (\sqrt{9}+1=\sqrt{10})
Because (9+1=\sqrt{10})
Because ((\sqrt{9})^2-1^2=10)
Hard · Level 6View options
\(\sqrt{1848}\)
\(\sqrt{1849}=43\)
\(\sqrt{1850}\)
\(\sqrt{1936}=44\)
Hard · Level 6View options
\(\sqrt{35}\) lies between 5 and 6, while \(\sqrt{37}\) lies between 6 and 7
Both lie between 5 and 6
Both lie between 6 and 7
\(\sqrt{35}\) lies between 6 and 7, while \(\sqrt{37}\) lies between 5 and 6
Hard · Level 6View options
\(\sqrt{2209}=47\)
\(\sqrt{2209}=46\)
\(\sqrt{2210}\)
\(\sqrt{4416}\)
Hard · Level 6View options
\(\sqrt{48}\) lies between 6 and 7, while \(\sqrt{50}\) lies between 7 and 8
Both lie between 7 and 8
Both lie between 6 and 7
Both are equal to 7
Hard · Level 6View options
√9999 lies between 99 and 100, and √10000 = 100
Both are 100
√9999 = 100 and √10000 is irrational
Both are greater than 100
Hard · Level 6View options
\((\sqrt{72})^2+1^2=73\), so the new hypotenuse is \(\sqrt{73}\)
\(\sqrt{72}+1\), so the new hypotenuse is \(\sqrt{72}+1\)
\(72^2+1^2=5185\), so the new hypotenuse is \(\sqrt{5185}\)
\((\sqrt{72})^2-1^2=71\), so the new hypotenuse is \(\sqrt{71}\)
Hard · Level 6View options
\(\sqrt{2601}=51\)
\(\sqrt{2599}\), जो 51 से कम है
\(\sqrt{5200}=10\sqrt{52}\)
\(\sqrt{2602}\), जो 51 से अधिक है
Hard · Level 6View options
The next hypotenuse is (\sqrt{16}=4), and (\sqrt{17}) lies between (4) and (5)
The next hypotenuse is (\sqrt{14}), and (\sqrt{17}=4)
Both are exactly at (4)
The next hypotenuse is (\sqrt{30}), and (\sqrt{17}) lies between (3) and (4)
Hard · Level 6View options
(57<\sqrt{3480}<58)
(58<\sqrt{3480}<59)
(59<\sqrt{3480}<60)
(60<\sqrt{3480}<61)
Hard · Level 6View options
\(\sqrt{3025}=55\)
\(\sqrt{3023}\), no whole value
\(\sqrt{6048}\), no whole value
\(\sqrt{3026}=55\)
Hard · Level 6View options
\(\sqrt{2207}\) lies between 46 and 47, and \(\sqrt{2209}=47\)
\(\sqrt{2207}=47\), and \(\sqrt{2209}\) is irrational
Both lie between 47 and 48
Both are exactly at 47
Hard · Level 6View options
√(n+8)
√(n+16)
√(n+64)
√(8n)
Hard · Level 6View options
(\sqrt{4094})
(\sqrt{4095})
(\sqrt{4096})
(\sqrt{4097})
Hard · Level 6View options
\(\sqrt{4096}=64\), next hypotenuse \(\sqrt{4097}\)
\(\sqrt{4096}=63\), next hypotenuse \(\sqrt{4097}\)
\(\sqrt{4096}=64\), next hypotenuse \(\sqrt{8192}\)
\(\sqrt{4096}\) is irrational, next hypotenuse \(\sqrt{4097}\)
Hard · Level 6View options
\(\sqrt{3598}\), (60)
\(\sqrt{3599}\), (60)
\(\sqrt{3601}\), (60)
\(\sqrt{3599}\), (59)
Hard · Level 6View options
The conclusion is correct, because \(64<70<81\).
The conclusion is incorrect; \(\sqrt{70}\) lies between 7 and 8.
The conclusion is correct, because 70 is a perfect square.
The conclusion is incorrect; \(\sqrt{70}\) lies between 9 and 10.
Hard · Level 6View options
\(\sqrt{1599}\)
\(\sqrt{1600}\)
\(\sqrt{1601}\)
\(\sqrt{1598}\)
Hard · Level 6View options
\(41<\sqrt{2023}<42\)
\(42<\sqrt{2023}<43\)
\(43<\sqrt{2023}<44\)
\(44<\sqrt{2023}<45\)
Hard · Level 6View options
(\sqrt{98^2+1^2}=\sqrt{99})
(\sqrt{98+2}=\sqrt{99})
(\sqrt{(\sqrt{98})^2+1^2}=\sqrt{99})
(\sqrt{98-1}=\sqrt{99})
Question 1HardLevel 6
In a square root spiral, which hypotenuse is formed by drawing a (1) unit perpendicular on \(\sqrt{1599}\)?
Correct answer: B
In a square root spiral, when one side is \(\sqrt{n}\) and the perpendicular side is 1 unit, the square of the new hypotenuse is \(n+1\). Therefore, drawing a 1-unit perpendicular on \(\sqrt{1599}\) gives \(\sqrt{1599+1}=\sqrt{1600}\). \(\sqrt{1598}\) would result from subtracting 1, while \(\sqrt{1601}\) would require an increase of 2. Exam tip: at each new step, add 1 to the number inside the square root.
While identifying the interval of \(\sqrt{1368}\) in a square root spiral, which conclusion is correct?
Correct answer: B
We have \(36^2=1296\) and \(37^2=1369\). Since \(1296<1368<1369\), taking square roots gives \(36<\sqrt{1368}<37\). It cannot equal 37 because \(37^2=1369\), not 1368. Exam tip: To locate a square root, compare the number with the nearest perfect squares on either side.
Which statement is correct when comparing \(\sqrt{960}\) and \(\sqrt{1024}\) in a square root spiral?
Correct answer: A
Since \(30^2=900\) and \(31^2=961\), and \(900<960<961\), we get \(30<\sqrt{960}<31\). Also, \(1024=32^2\), so \(\sqrt{1024}=32\). Option B is incorrect because 960 is less than \(31^2\), so its square root cannot be greater than 31. Exam tip: Compare a number with nearby perfect squares to estimate its square root.
Why is using (\sqrt{9}) and (1) correct for constructing (\sqrt{10}) in a square root spiral?
Correct answer: A
In a square root spiral, the existing hypotenuse is used as one leg of a new right triangle, and a perpendicular segment of length 1 is added. If the existing hypotenuse is \(\sqrt{9}\), then the square of the next hypotenuse is \((\sqrt{9})^2+1^2=9+1=10\). Thus the new hypotenuse is \(\sqrt{10}\). The construction depends on adding squares, not adding the lengths themselves.
Therefore option A is correct. Option B incorrectly claims \(\sqrt{9}+1=\sqrt{10}\); in fact, \(3+1=4\), not \(\sqrt{10}\). Option C confuses the number 10 with its square root, while option D subtracts the squares and does not represent the right-triangle construction. The previous hypotenuse and the unit perpendicular work together through the Pythagorean theorem, so the supplied answer is accurate.
In a square root spiral, the hypotenuse formed after \(\sqrt{1848}\) will be located at which special value?
Correct answer: B
In a square root spiral, successive hypotenuses are \(\sqrt{1},\sqrt{2},\sqrt{3}\), and so on. Therefore, the hypotenuse after \(\sqrt{1848}\) is \(\sqrt{1849}\). Since \(1849=43^2\), we get \(\sqrt{1849}=43\), so it lies at an integer value. \(\sqrt{1850}\) comes next, but 1850 is not a perfect square. Exam tip: add 1 to the radicand for the next hypotenuse, then check whether it is a perfect square.
Which statement about the positions of \(\sqrt{35}\) and \(\sqrt{37}\) in a square root spiral is correct?
Correct answer: A
Since \(5^2=25<35<36=6^2\), we get \(5<\sqrt{35}<6\). Similarly, \(6^2=36<37<49=7^2\), so \(6<\sqrt{37}<7\). Therefore, option A is correct. Option B places \(\sqrt{37}\) incorrectly, while option C places \(\sqrt{35}\) incorrectly. Exam tip: compare a number with consecutive perfect squares to locate its square root.
In a square root spiral, which hypotenuse is formed after \(\sqrt{2208}\), and what is its exact value?
Correct answer: A
In a square root spiral, successive hypotenuses are \(\sqrt{1}, \sqrt{2}, \sqrt{3}\), and so on. Therefore, the hypotenuse after \(\sqrt{2208}\) is \(\sqrt{2209}\). Since \(47^2=2209\), \(\sqrt{2209}=47\). Option B has the correct radicand but an incorrect value, while option C skips to the next term. Exam tip: identify nearby perfect squares to check whether a square root is an integer.
Which statement is correct when comparing \(\sqrt{48}\) and \(\sqrt{50}\) in a square root spiral?
Correct answer: A
Since \(6^2=36\), \(7^2=49\), and \(8^2=64\), \(36<48<49\) gives \(6<\sqrt{48}<7\). Similarly, \(49<50<64\) gives \(7<\sqrt{50}<8\). Therefore, option A is correct. A common confusion is that both numbers are close to 49, but 48 is below 49 whereas 50 is above it. Exam tip: locate a square root by comparing the number with the nearest perfect squares.
Which is the correct comparison of √9999 and √10000 in a square root spiral?
Correct answer: A
The correct answer is A. Compare the radicand with consecutive squares: 99² = 9801 and 100² = 10000. Since 9801 < 9999 < 10000, taking positive square roots gives 99 < √9999 < 100. Also, 10000 = 100², so √10000 = 100 exactly. The two values are therefore not equal: √9999 is slightly smaller than 100, while √10000 reaches the integer 100. Option B ignores the strict inequality, option C reverses the perfect-square facts and misclassifies √10000, and option D contradicts the upper bound obtained from 100².
Which option is logical for finding the next hypotenuse from \(\sqrt{72}\) in a square root spiral?
Correct answer: A
In a square root spiral, each new right triangle has the previous hypotenuse as one side and a unit length as the other side. By Pythagoras’ theorem, the next hypotenuse is \(\sqrt{(\sqrt{72})^2+1^2}=\sqrt{73}\). Option B incorrectly adds the hypotenuses directly; in general, the sum of square roots is not the square root of their sum. Exam tip: for the next hypotenuse, add \(1\) to the square of the previous hypotenuse.
In a square root spiral, which hypotenuse is formed after \(\sqrt{2600}\), and what is its exact value?
Correct answer: A
In a square root spiral, successive hypotenuses are \(\sqrt{1}, \sqrt{2}, \sqrt{3}\), and so on. Therefore, the hypotenuse after \(\sqrt{2600}\) is \(\sqrt{2601}\). Since \(2601=51^2\), we get \(\sqrt{2601}=51\). \(\sqrt{2602}\) is the hypotenuse after the next one, not the immediate next hypotenuse. Exam tip: For a number near a perfect square, first check whether it equals the square of an integer.
If a (1) unit perpendicular is drawn on hypotenuse \(\sqrt{3024}\) in a square root spiral, what will be the new hypotenuse and its exact value?
Correct answer: A
In a square-root spiral, adding a perpendicular segment of length 1 creates a right triangle. If the old hypotenuse is sqrt{3024} , the new hypotenuse is found by adding the square of the new unit side: sqrt{( sqrt{3024})^2+1^2}= sqrt{3024+1}= sqrt{3025} . Since 3025=55^2 , its principal square root is 55. The positive value is used because a length cannot be negative.
Therefore option A is correct. Option B subtracts 1 instead of adding the perpendicular side's square. Option C doubles 3024, which does not follow from the theorem, and option D adds 2 rather than 1 while also claiming the incorrect equality sqrt{3026}=55 . The exact radical form is sqrt{3025} , and its simplified exact value is 55.
Which statement about the number-line positions of \(\sqrt{2207}\) and \(\sqrt{2209}\) in a square root spiral is correct?
Correct answer: A
Since \(46^2=2116\), \(47^2=2209\), and \(2116<2207<2209\), we get \(46<\sqrt{2207}<47\). Also, \(2209=47^2\), so \(\sqrt{2209}=47\). Option D may seem close, but \(2207\) is less than \(2209\), so its square root cannot be 47. Exam tip: locate a square root by comparing the number with nearby perfect squares.
If an 8-unit perpendicular is drawn on √n in the usual square root spiral setup, what form will the formed hypotenuse have?
Correct answer: C
The relevant concept is the Pythagorean theorem for a right triangle. One perpendicular side has length √n and the other has length 8. If H is the new hypotenuse, then H² = (√n)² + 8² = n + 64. Since a geometric length is positive, H = √(n+64). Thus option C is correct. Option A adds 8 instead of its square, so it ignores the theorem. Option B would correspond to adding 4² rather than 8². Option D multiplies n by 8 and does not represent the sum of the squares of the perpendicular sides. The essential step is recognizing that the added side contributes 8²=64 to the radicand.
In a square root spiral, what is the exact value of \(\sqrt{4096}\), and what is the form of the next hypotenuse?
Correct answer: A
Since \(4096=64^2\), \(\sqrt{4096}=64\) is an integer, not an irrational number. In a square root spiral, each new step adds \(1\) to the square of the previous hypotenuse. Therefore, the hypotenuse after \(\sqrt{4096}\) is \(\sqrt{4096+1}=\sqrt{4097}\), not \(\sqrt{8192}\). Exam tip: first check whether the radicand is a perfect square, then add \(1\) to the radicand for the next hypotenuse.
While representing \(\sqrt{70}\) on the number line using a square root spiral, Neelam places it between 8 and 9. What is the correct evaluation of her conclusion?
Correct answer: A
Neelam’s conclusion is correct. We have \(8^2=64\) and \(9^2=81\). Since \(64<70<81\), taking square roots gives \(8<\sqrt{70}<9\). Option B is incorrect because 70 does not lie between \(7^2=49\) and \(8^2=64\). Exam tip: locate a square root by comparing the number with consecutive perfect squares.
If the new hypotenuse in a square root spiral is equal to (40), which hypotenuse came immediately before it?
Correct answer: A
In a square root spiral, the successive hypotenuses are \(\sqrt{2}, \sqrt{3}, \sqrt{4}, \ldots\). Since the new hypotenuse is \(40=\sqrt{1600}\), the hypotenuse immediately before it is \(\sqrt{1599}\). \(\sqrt{1601}\) would be the next hypotenuse, while \(\sqrt{1598}\) is one step earlier. Exam tip: To place a whole number in this sequence, write it as \(\sqrt{n^2}\).
Before placing \(\sqrt{2023}\) on the number line, which interval will be correctly identified?
Correct answer: D
We have \(44^2=1936\) and \(45^2=2025\). Since \(1936<2023<2025\), taking positive square roots gives \(44<\sqrt{2023}<45\). The interval between 43 and 44 is not correct because \(44^2\) is still less than 2023. Exam tip: To locate a square root, compare the number with consecutive perfect squares.
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