The correct hypotenuse is (\sqrt{\(\sqrt{18}\)2+12}=\sqrt{19}). In a square root spiral, the number increases by (1) at each step.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{18+1}\). The correct hypotenuse is (\sqrt{\(\sqrt{18}\)2+12}=\sqrt{19}). In a square root spiral, the number increases by (1) at each step.
Step 3
Exam Tip
सही कर्ण (\sqrt{\(\sqrt{18}\)2+12}=\sqrt{19}) होगा। वर्गमूल सर्पिल में हर चरण में संख्या (1) बढ़ती है।
Since (\(\sqrt{27}\)2+12=28). Therefore \(\sqrt{27}\) is the correct previous hypotenuse.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{27}\) और (1) / \(\sqrt{27}\) and (1). Since (\(\sqrt{27}\)2+12=28). Therefore \(\sqrt{27}\) is the correct previous hypotenuse.
Step 3
Exam Tip
(\(\sqrt{27}\)2+12=28) होता है। इसलिए पिछले कर्ण के रूप में \(\sqrt{27}\) सही है।
The new hypotenuse is \(\sqrt{323+1}=\sqrt{324}\), and \(\sqrt{324}=18\). When a perfect square appears, write the exact whole number.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{324}\), (18). The new hypotenuse is \(\sqrt{323+1}=\sqrt{324}\), and \(\sqrt{324}=18\). When a perfect square appears, write the exact whole number.
Step 3
Exam Tip
नया कर्ण \(\sqrt{323+1}=\sqrt{324}\) होगा और \(\sqrt{324}=18\) है। पूर्ण वर्ग मिलने पर सटीक पूर्ण संख्या लिखें।
If the new hypotenuse is \(\sqrt{n+1}\), the previous hypotenuse is \(\sqrt{n}\). Therefore before \(\sqrt{35}\), it is \(\sqrt{34}\).
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{34}\). If the new hypotenuse is \(\sqrt{n+1}\), the previous hypotenuse is \(\sqrt{n}\). Therefore before \(\sqrt{35}\), it is \(\sqrt{34}\).
Step 3
Exam Tip
नया कर्ण \(\sqrt{n+1}\) हो तो पिछला कर्ण \(\sqrt{n}\) होता है। इसलिए \(\sqrt{35}\) से पहले \(\sqrt{34}\) होगा।
The next hypotenuse will be \(\sqrt{100}\), and \(\sqrt{100}=10\). Therefore it is a whole number.
Step 2
Why this answer is correct
The correct answer is A. पूर्ण संख्या / Whole number. The next hypotenuse will be \(\sqrt{100}\), and \(\sqrt{100}=10\). Therefore it is a whole number.
Step 3
Exam Tip
अगला कर्ण \(\sqrt{100}\) होगा और \(\sqrt{100}=10\) है। इसलिए वह पूर्ण संख्या है।
A. क्योंकि नया कर्ण (\sqrt{\(\sqrt{6}\)2+12}) से बनता है/Because the new hypotenuse is formed by (\sqrt{\(\sqrt{6}\)2+12})
Step 1
Concept
In a square root spiral, lengths are not added directly. The correct method is Pythagoras theorem.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि नया कर्ण (\sqrt{\(\sqrt{6}\)2+12}) से बनता है / Because the new hypotenuse is formed by (\sqrt{\(\sqrt{6}\)2+12}). In a square root spiral, lengths are not added directly. The correct method is Pythagoras theorem.
Step 3
Exam Tip
वर्गमूल सर्पिल में सीधे लंबाइयाँ नहीं जोड़ी जातीं। सही तरीका पाइथागोरस प्रमेय है।
B. \(\sqrt{72}\) (8) और (9) के बीच है और \(\sqrt{81}=9\) है/\(\sqrt{72}\) lies between (8) and (9), and \(\sqrt{81}=9\)
Step 1
Concept
\(8^2<72<9^2\), and \(81=9^2\). Therefore \(\sqrt{81}\) equals (9).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{72}\) (8) और (9) के बीच है और \(\sqrt{81}=9\) है / \(\sqrt{72}\) lies between (8) and (9), and \(\sqrt{81}=9\). \(8^2<72<9^2\), and \(81=9^2\). Therefore \(\sqrt{81}\) equals (9).
Step 3
Exam Tip
\(8^2<72<9^2\) और \(81=9^2\) है। इसलिए \(\sqrt{81}\) का मान (9) है।
\(\sqrt{120}\) is formed from \(\sqrt{119}\) and a (1) unit perpendicular. The previous hypotenuse number is always (1) less.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{119}\). \(\sqrt{120}\) is formed from \(\sqrt{119}\) and a (1) unit perpendicular. The previous hypotenuse number is always (1) less.
Step 3
Exam Tip
\(\sqrt{119}\) और (1) इकाई लंब से \(\sqrt{120}\) बनता है। पिछले कर्ण की संख्या हमेशा (1) कम होती है।
A. क्योंकि \(1^2\) की जगह \(4^2\) जुड़ जाएगा/Because \(4^2\) will be added instead of \(1^2\)
Step 1
Concept
In the usual spiral \(1^2\) is added every time. Taking (4) units adds (16), so the sequence changes.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि \(1^2\) की जगह \(4^2\) जुड़ जाएगा / Because \(4^2\) will be added instead of \(1^2\). In the usual spiral \(1^2\) is added every time. Taking (4) units adds (16), so the sequence changes.
Step 3
Exam Tip
सामान्य सर्पिल में हर बार \(1^2\) जुड़ता है। (4) इकाई लेने पर (16) जुड़ता है और क्रम बदल जाता है।
C. पिछले कर्ण के वर्ग में \(1^2\) जुड़ता है/\(1^2\) is added to the square of the previous hypotenuse
Step 1
Concept
By Pythagoras theorem, (\(\sqrt{14}\)2+12=15). Therefore the new hypotenuse is \(\sqrt{15}\).
Step 2
Why this answer is correct
The correct answer is C. पिछले कर्ण के वर्ग में \(1^2\) जुड़ता है / \(1^2\) is added to the square of the previous hypotenuse. By Pythagoras theorem, (\(\sqrt{14}\)2+12=15). Therefore the new hypotenuse is \(\sqrt{15}\).
Step 3
Exam Tip
पाइथागोरस प्रमेय से (\(\sqrt{14}\)2+12=15) होता है। इसलिए नया कर्ण \(\sqrt{15}\) बनता है।
B. \(\sqrt{7}\) और (1) भुजाओं वाला/With sides \(\sqrt{7}\) and (1)
Step 1
Concept
(\(\sqrt{7}\)2+12=8). So \(\sqrt{7}\) and (1) are correct to form hypotenuse \(\sqrt{8}\).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{7}\) और (1) भुजाओं वाला / With sides \(\sqrt{7}\) and (1). (\(\sqrt{7}\)2+12=8). So \(\sqrt{7}\) and (1) are correct to form hypotenuse \(\sqrt{8}\).
Step 3
Exam Tip
(\(\sqrt{7}\)2+12=8) होता है। इसलिए कर्ण \(\sqrt{8}\) बनाने के लिए \(\sqrt{7}\) और (1) सही हैं।
A. दोनों (5) और (6) के बीच हैं/Both are between (5) and (6)
Step 1
Concept
Both \(5^2<26<6^2\) and \(5^2<27<6^2\) are true. Therefore both lie between (5) and (6).
Step 2
Why this answer is correct
The correct answer is A. दोनों (5) और (6) के बीच हैं / Both are between (5) and (6). Both \(5^2<26<6^2\) and \(5^2<27<6^2\) are true. Therefore both lie between (5) and (6).
Step 3
Exam Tip
\(5^2<26<6^2\) और \(5^2<27<6^2\) दोनों सही हैं। इसलिए दोनों (5) और (6) के बीच हैं।
A. \(\sqrt{144}\), ठीक (12) पर/\(\sqrt{144}\), exactly at (12)
Step 1
Concept
The next hypotenuse is \(\sqrt{144}\), and \(\sqrt{144}=12\). It is not in an interval, but exactly at (12).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{144}\), ठीक (12) पर / \(\sqrt{144}\), exactly at (12). The next hypotenuse is \(\sqrt{144}\), and \(\sqrt{144}=12\). It is not in an interval, but exactly at (12).
Step 3
Exam Tip
अगला कर्ण \(\sqrt{144}\) है और \(\sqrt{144}=12\) होता है। यह अंतराल में नहीं, ठीक (12) पर है।
A. कंपास, कर्ण लंबाई को संख्या रेखा पर स्थानांतरित करने के लिए/Compass, to transfer hypotenuse length to the number line
Step 1
Concept
A compass is used to take the hypotenuse length and draw an arc on the number line. This is the correct geometric transfer method.
Step 2
Why this answer is correct
The correct answer is A. कंपास, कर्ण लंबाई को संख्या रेखा पर स्थानांतरित करने के लिए / Compass, to transfer hypotenuse length to the number line. A compass is used to take the hypotenuse length and draw an arc on the number line. This is the correct geometric transfer method.
Step 3
Exam Tip
कंपास से कर्ण की लंबाई लेकर संख्या रेखा पर चाप खींचा जाता है। यह ज्यामितीय स्थानांतरण का सही तरीका है।
B. अपरिमेय संख्या और (7) तथा (8) के बीच/Irrational number and between (7) and (8)
Step 1
Concept
(55) is not a perfect square and \(7^2<55<8^2\). Therefore \(\sqrt{55}\) is irrational and lies between (7) and (8).
Step 2
Why this answer is correct
The correct answer is B. अपरिमेय संख्या और (7) तथा (8) के बीच / Irrational number and between (7) and (8). (55) is not a perfect square and \(7^2<55<8^2\). Therefore \(\sqrt{55}\) is irrational and lies between (7) and (8).
Step 3
Exam Tip
(55) पूर्ण वर्ग नहीं है और \(7^2<55<8^2\) है। इसलिए \(\sqrt{55}\) अपरिमेय है और (7) तथा (8) के बीच है।
B. \(\sqrt{1}\rightarrow\sqrt{2}\rightarrow\sqrt{3}\rightarrow\sqrt{4}\)
Step 1
Concept
In a square root spiral, hypotenuses are formed in order. In the usual construction, the \(\sqrt{3}\) step is not skipped.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{1}\rightarrow\sqrt{2}\rightarrow\sqrt{3}\rightarrow\sqrt{4}\). In a square root spiral, hypotenuses are formed in order. In the usual construction, the \(\sqrt{3}\) step is not skipped.
Step 3
Exam Tip
वर्गमूल सर्पिल में कर्ण क्रम से बनते हैं। सामान्य निर्माण में \(\sqrt{3}\) चरण छोड़ा नहीं जाता।
A. दोनों में समकोण त्रिभुज और पाइथागोरस प्रमेय का उपयोग होता है/Both use a right triangle and Pythagoras theorem
Step 1
Concept
Both constructions are based on right triangles. The hypotenuse is found by Pythagoras theorem.
Step 2
Why this answer is correct
The correct answer is A. दोनों में समकोण त्रिभुज और पाइथागोरस प्रमेय का उपयोग होता है / Both use a right triangle and Pythagoras theorem. Both constructions are based on right triangles. The hypotenuse is found by Pythagoras theorem.
Step 3
Exam Tip
दोनों निर्माण समकोण त्रिभुज पर आधारित हैं। कर्ण पाइथागोरस प्रमेय से निकाला जाता है।
B. \(\sqrt{101}\) (10) और (11) के बीच है और \(\sqrt{121}=11\) है/\(\sqrt{101}\) is between (10) and (11), and \(\sqrt{121}=11\)
Step 1
Concept
\(10^2<101<11^2\), and \(121=11^2\). Therefore \(\sqrt{121}=11\).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{101}\) (10) और (11) के बीच है और \(\sqrt{121}=11\) है / \(\sqrt{101}\) is between (10) and (11), and \(\sqrt{121}=11\). \(10^2<101<11^2\), and \(121=11^2\). Therefore \(\sqrt{121}=11\).
Step 3
Exam Tip
\(10^2<101<11^2\) और \(121=11^2\) है। इसलिए \(\sqrt{121}=11\) है।
\(\sqrt{59}\) is the previous hypotenuse and the new perpendicular is (1) unit. Therefore the new hypotenuse is \(\sqrt{60}\).
Step 2
Why this answer is correct
The correct answer is A. (\(\sqrt{59}\)2+12=60). \(\sqrt{59}\) is the previous hypotenuse and the new perpendicular is (1) unit. Therefore the new hypotenuse is \(\sqrt{60}\).
Step 3
Exam Tip
\(\sqrt{59}\) पिछले कर्ण की लंबाई है और नई लंब (1) इकाई है। इसलिए नया कर्ण \(\sqrt{60}\) होगा।
A whole number square root is obtained only when (n) is a perfect square. For example, \(\sqrt{81}=9\).
Step 2
Why this answer is correct
The correct answer is A. (n) पूर्ण वर्ग है / (n) is a perfect square. A whole number square root is obtained only when (n) is a perfect square. For example, \(\sqrt{81}=9\).
Step 3
Exam Tip
पूर्ण संख्या वर्गमूल तभी मिलता है जब (n) पूर्ण वर्ग हो। जैसे \(\sqrt{81}=9\)।
B. (\(\sqrt{3}\)2+12=4) इसलिए कर्ण \(\sqrt{4}\) है/(\(\sqrt{3}\)2+12=4), so the hypotenuse is \(\sqrt{4}\)
Step 1
Concept
In a square root spiral, the sum of squares is taken by Pythagoras theorem. Directly adding lengths is wrong.
Step 2
Why this answer is correct
The correct answer is B. (\(\sqrt{3}\)2+12=4) इसलिए कर्ण \(\sqrt{4}\) है / (\(\sqrt{3}\)2+12=4), so the hypotenuse is \(\sqrt{4}\). In a square root spiral, the sum of squares is taken by Pythagoras theorem. Directly adding lengths is wrong.
Step 3
Exam Tip
वर्गमूल सर्पिल में पाइथागोरस प्रमेय से वर्गों का योग लिया जाता है। सीधे लंबाइयाँ जोड़ना गलत है।
B. दोनों अपरिमेय संख्याएँ हैं/Both are irrational numbers
Step 1
Concept
The numbers (2) and (8) are not perfect squares. Therefore both square roots are irrational.
Step 2
Why this answer is correct
The correct answer is B. दोनों अपरिमेय संख्याएँ हैं / Both are irrational numbers. The numbers (2) and (8) are not perfect squares. Therefore both square roots are irrational.
Step 3
Exam Tip
(2) और (8) पूर्ण वर्ग नहीं हैं। इसलिए दोनों के वर्गमूल अपरिमेय हैं।
(\(\sqrt{4}\)2+12=5). Therefore sides \(\sqrt{4}\) and (1) form hypotenuse \(\sqrt{5}\).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{4}\) और (1) / \(\sqrt{4}\) and (1). (\(\sqrt{4}\)2+12=5). Therefore sides \(\sqrt{4}\) and (1) form hypotenuse \(\sqrt{5}\).
Step 3
Exam Tip
(\(\sqrt{4}\)2+12=5) है। इसलिए \(\sqrt{4}\) और (1) से कर्ण \(\sqrt{5}\) बनेगा।
The next hypotenuse will be \(\sqrt{64}\), and \(\sqrt{64}=8\). At a perfect square, an exact whole number is obtained.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{64}\), (8). The next hypotenuse will be \(\sqrt{64}\), and \(\sqrt{64}=8\). At a perfect square, an exact whole number is obtained.
Step 3
Exam Tip
अगला कर्ण \(\sqrt{64}\) होगा और \(\sqrt{64}=8\) है। पूर्ण वर्ग पर सटीक पूर्ण संख्या मिलती है।
A. \(\sqrt{90}\) (9) और (10) के बीच है और \(\sqrt{100}=10\) है/\(\sqrt{90}\) is between (9) and (10), and \(\sqrt{100}=10\)
Step 1
Concept
\(9^2<90<10^2\), and \(100=10^2\). Therefore the first statement is correct.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{90}\) (9) और (10) के बीच है और \(\sqrt{100}=10\) है / \(\sqrt{90}\) is between (9) and (10), and \(\sqrt{100}=10\). \(9^2<90<10^2\), and \(100=10^2\). Therefore the first statement is correct.
Step 3
Exam Tip
\(9^2<90<10^2\) और \(100=10^2\) है। इसलिए तुलना में पहला कथन सही है।
A. इनसे (\(\sqrt{n}\)2+12=n+1) लागू होता है/They allow (\(\sqrt{n}\)2+12=n+1) to apply
Step 1
Concept
The (1) unit perpendicular and right angle allow Pythagoras theorem to apply correctly. This forms successive square roots.
Step 2
Why this answer is correct
The correct answer is A. इनसे (\(\sqrt{n}\)2+12=n+1) लागू होता है / They allow (\(\sqrt{n}\)2+12=n+1) to apply. The (1) unit perpendicular and right angle allow Pythagoras theorem to apply correctly. This forms successive square roots.
Step 3
Exam Tip
(1) इकाई लंब और समकोण से पाइथागोरस प्रमेय सही लागू होता है। इसी से क्रमिक वर्गमूल बनते हैं।
D. अगला कर्ण पिछले कर्ण में सीधे (1) जोड़कर निकालना/Finding the next hypotenuse by directly adding (1) to the previous hypotenuse
Step 1
Concept
Direct addition is not used in a square root spiral. The new hypotenuse is obtained from a right triangle and Pythagoras theorem.
Step 2
Why this answer is correct
The correct answer is D. अगला कर्ण पिछले कर्ण में सीधे (1) जोड़कर निकालना / Finding the next hypotenuse by directly adding (1) to the previous hypotenuse. Direct addition is not used in a square root spiral. The new hypotenuse is obtained from a right triangle and Pythagoras theorem.
Step 3
Exam Tip
वर्गमूल सर्पिल में सीधे जोड़ नहीं किया जाता। नया कर्ण समकोण त्रिभुज और पाइथागोरस प्रमेय से मिलता है।
Adding a (1) unit perpendicular to \(\sqrt{223}\) forms \(\sqrt{224}\). The number increases by (1) in the next hypotenuse.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{223}\). Adding a (1) unit perpendicular to \(\sqrt{223}\) forms \(\sqrt{224}\). The number increases by (1) in the next hypotenuse.
Step 3
Exam Tip
\(\sqrt{223}\) में (1) इकाई लंब जोड़ने पर \(\sqrt{224}\) बनता है। अगले कर्ण में संख्या (1) बढ़ती है।
A. \(\sqrt{2}\) कर्ण की लंबाई कंपास में लेकर मूल बिंदु से चाप खींचना/Take the \(\sqrt{2}\) hypotenuse length in a compass and draw an arc from the origin
Step 1
Concept
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.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}\) कर्ण की लंबाई कंपास में लेकर मूल बिंदु से चाप खींचना / Take the \(\sqrt{2}\) hypotenuse length in a compass and draw an arc from the origin. 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.
Step 3
Exam Tip
जिस वर्गमूल को अंकित करना है, उसी कर्ण की लंबाई कंपास में ली जाती है। मूल बिंदु से चाप खींचना सही तरीका है।
A. पिछला कर्ण \(\sqrt{n}\) और नई लंब (1) इकाई/Previous hypotenuse \(\sqrt{n}\) and new perpendicular (1) unit
Step 1
Concept
Because (\(\sqrt{n}\)2+12=n+1). Therefore the new hypotenuse becomes \(\sqrt{n+1}\).
Step 2
Why this answer is correct
The correct answer is A. पिछला कर्ण \(\sqrt{n}\) और नई लंब (1) इकाई / Previous hypotenuse \(\sqrt{n}\) and new perpendicular (1) unit. Because (\(\sqrt{n}\)2+12=n+1). Therefore the new hypotenuse becomes \(\sqrt{n+1}\).
Step 3
Exam Tip
क्योंकि (\(\sqrt{n}\)2+12=n+1) होता है। इसलिए नया कर्ण \(\sqrt{n+1}\) बनता है।
A. \(\sqrt{24}\) (4) और (5) के बीच है और \(\sqrt{25}=5\) है/\(\sqrt{24}\) is between (4) and (5), and \(\sqrt{25}=5\)
Step 1
Concept
\(4^2<24<5^2\), and \(25=5^2\). Therefore the exact value of \(\sqrt{25}\) is (5).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{24}\) (4) और (5) के बीच है और \(\sqrt{25}=5\) है / \(\sqrt{24}\) is between (4) and (5), and \(\sqrt{25}=5\). \(4^2<24<5^2\), and \(25=5^2\). Therefore the exact value of \(\sqrt{25}\) is (5).
Step 3
Exam Tip
\(4^2<24<5^2\) और \(25=5^2\) है। इसलिए \(\sqrt{25}\) का सटीक मान (5) है।
A. नया कर्ण पूर्ण संख्या (4) बन जाता है/The new hypotenuse becomes the whole number (4)
Step 1
Concept
\(\sqrt{16}=4\). When the number is a perfect square, the square root becomes a whole number.
Step 2
Why this answer is correct
The correct answer is A. नया कर्ण पूर्ण संख्या (4) बन जाता है / The new hypotenuse becomes the whole number (4). \(\sqrt{16}=4\). When the number is a perfect square, the square root becomes a whole number.
Step 3
Exam Tip
\(\sqrt{16}=4\) है। जब संख्या पूर्ण वर्ग हो, तो वर्गमूल पूर्ण संख्या बनता है।
Since (\(\sqrt{109}\)2+12=110). Therefore \(\sqrt{109}\) is the correct previous hypotenuse.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{109}\) और (1) / \(\sqrt{109}\) and (1). Since (\(\sqrt{109}\)2+12=110). Therefore \(\sqrt{109}\) is the correct previous hypotenuse.
Step 3
Exam Tip
(\(\sqrt{109}\)2+12=110) होता है। इसलिए पिछले कर्ण के रूप में \(\sqrt{109}\) सही है।
A. \(\sqrt{2}\rightarrow\sqrt{3}\rightarrow\sqrt{4}\rightarrow\sqrt{5}\)
Step 1
Concept
In a square root spiral, hypotenuses are formed successively. In the usual construction, intermediate steps are not skipped.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}\rightarrow\sqrt{3}\rightarrow\sqrt{4}\rightarrow\sqrt{5}\). In a square root spiral, hypotenuses are formed successively. In the usual construction, intermediate steps are not skipped.
Step 3
Exam Tip
वर्गमूल सर्पिल में कर्ण क्रमिक रूप से बनते हैं। सामान्य निर्माण में बीच के चरण छोड़े नहीं जाते।
A. यह समकोण त्रिभुजों की श्रृंखला है, जहाँ पिछला कर्ण और (1) इकाई लंब अगला वर्गमूल बनाते हैं/It is a chain of right triangles where the previous hypotenuse and (1) unit perpendicular form the next square root
Step 1
Concept
A square root spiral is a stepwise construction based on Pythagoras theorem. Each new hypotenuse gives the next square root.
Step 2
Why this answer is correct
The correct answer is A. यह समकोण त्रिभुजों की श्रृंखला है, जहाँ पिछला कर्ण और (1) इकाई लंब अगला वर्गमूल बनाते हैं / It is a chain of right triangles where the previous hypotenuse and (1) unit perpendicular form the next square root. A square root spiral is a stepwise construction based on Pythagoras theorem. Each new hypotenuse gives the next square root.
Step 3
Exam Tip
वर्गमूल सर्पिल पाइथागोरस प्रमेय पर आधारित क्रमिक निर्माण है। हर नया कर्ण अगला वर्गमूल देता है।
