In a square root spiral, the hypotenuse is formed by the sum of squares. Do not treat \(\sqrt{18}+1\) as \(\sqrt{19}\).
Step 2
Why this answer is correct
The correct answer is A. (\sqrt{\(\sqrt{18}\)2+12}=\sqrt{19}). In a square root spiral, the hypotenuse is formed by the sum of squares. Do not treat \(\sqrt{18}+1\) as \(\sqrt{19}\).
Step 3
Exam Tip
वर्गमूल सर्पिल में कर्ण वर्गों के योग से बनता है। \(\sqrt{18}+1\) को \(\sqrt{19}\) नहीं मानना चाहिए।
A. \(\sqrt{170}\), (13) और (14) के बीच/\(\sqrt{170}\), between (13) and (14)
Step 1
Concept
The new hypotenuse is \(\sqrt{170}\). Since \(13^2<170<14^2\), it lies between (13) and (14).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{170}\), (13) और (14) के बीच / \(\sqrt{170}\), between (13) and (14). The new hypotenuse is \(\sqrt{170}\). Since \(13^2<170<14^2\), it lies between (13) and (14).
Step 3
Exam Tip
नया कर्ण \(\sqrt{170}\) है। क्योंकि \(13^2<170<14^2\), यह (13) और (14) के बीच है।
B. \(\sqrt{224}\) (14) और (15) के बीच है और \(\sqrt{225}=15\) है/\(\sqrt{224}\) lies between (14) and (15), and \(\sqrt{225}=15\)
Step 1
Concept
\(14^2<224<15^2\), and \(225=15^2\). Therefore \(\sqrt{225}\) is exactly (15).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{224}\) (14) और (15) के बीच है और \(\sqrt{225}=15\) है / \(\sqrt{224}\) lies between (14) and (15), and \(\sqrt{225}=15\). \(14^2<224<15^2\), and \(225=15^2\). Therefore \(\sqrt{225}\) is exactly (15).
Step 3
Exam Tip
\(14^2<224<15^2\) और \(225=15^2\) है। इसलिए \(\sqrt{225}\) ठीक (15) है।
If the (k)-th hypotenuse is \(\sqrt{k}\), then \(\sqrt{36}\) is the (36)-th hypotenuse. Also, \(\sqrt{36}=6\).
Step 2
Why this answer is correct
The correct answer is B. (36)वाँ, (6) / (36)-th, (6). If the (k)-th hypotenuse is \(\sqrt{k}\), then \(\sqrt{36}\) is the (36)-th hypotenuse. Also, \(\sqrt{36}=6\).
Step 3
Exam Tip
यदि (k)वाँ कर्ण \(\sqrt{k}\) है, तो \(\sqrt{36}\) (36)वाँ कर्ण है। \(\sqrt{36}=6\) होता है।
(\(\sqrt{119}\)2+12=120). So the previous hypotenuse for \(\sqrt{120}\) is \(\sqrt{119}\).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{119}\) और (1) / \(\sqrt{119}\) and (1). (\(\sqrt{119}\)2+12=120). So the previous hypotenuse for \(\sqrt{120}\) is \(\sqrt{119}\).
Step 3
Exam Tip
(\(\sqrt{119}\)2+12=120) है। इसलिए \(\sqrt{120}\) के लिए पिछला कर्ण \(\sqrt{119}\) होगा।
B. \(\sqrt{2}\rightarrow\sqrt{3}\rightarrow\sqrt{4}\rightarrow\sqrt{5}\)
Step 1
Concept
Hypotenuses are formed successively in the spiral. In the usual construction, intermediate square roots are not skipped.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{2}\rightarrow\sqrt{3}\rightarrow\sqrt{4}\rightarrow\sqrt{5}\). Hypotenuses are formed successively in the spiral. In the usual construction, intermediate square roots are not skipped.
Step 3
Exam Tip
सर्पिल में कर्ण क्रमिक रूप से बनते हैं। सामान्य निर्माण में बीच के वर्गमूल नहीं छोड़े जाते।
A. दोनों (13) और (14) के बीच हैं/Both lie between (13) and (14)
Step 1
Concept
Because \(13^2<170<14^2\) and \(13^2<195<14^2\). Therefore both lie between (13) and (14).
Step 2
Why this answer is correct
The correct answer is A. दोनों (13) और (14) के बीच हैं / Both lie between (13) and (14). Because \(13^2<170<14^2\) and \(13^2<195<14^2\). Therefore both lie between (13) and (14).
Step 3
Exam Tip
क्योंकि \(13^2<170<14^2\) और \(13^2<195<14^2\) हैं। इसलिए दोनों (13) और (14) के बीच हैं।
A. अगला कर्ण \(\sqrt{25}=5\) है और \(\sqrt{26}\) (5) और (6) के बीच है/The next hypotenuse is \(\sqrt{25}=5\), and \(\sqrt{26}\) lies between (5) and (6)
Step 1
Concept
After \(\sqrt{24}\), \(\sqrt{25}=5\) is formed. Since \(5^2<26<6^2\), \(\sqrt{26}\) lies between (5) and (6).
Step 2
Why this answer is correct
The correct answer is A. अगला कर्ण \(\sqrt{25}=5\) है और \(\sqrt{26}\) (5) और (6) के बीच है / The next hypotenuse is \(\sqrt{25}=5\), and \(\sqrt{26}\) lies between (5) and (6). After \(\sqrt{24}\), \(\sqrt{25}=5\) is formed. Since \(5^2<26<6^2\), \(\sqrt{26}\) lies between (5) and (6).
Step 3
Exam Tip
\(\sqrt{24}\) के बाद \(\sqrt{25}=5\) बनता है। \(5^2<26<6^2\), इसलिए \(\sqrt{26}\) (5) और (6) के बीच है।
The next hypotenuse is \(\sqrt{m+1}\). If (m+1) is a perfect square, its square root will be a whole number.
Step 2
Why this answer is correct
The correct answer is A. वह पूर्ण संख्या होगा / It will be a whole number. The next hypotenuse is \(\sqrt{m+1}\). If (m+1) is a perfect square, its square root will be a whole number.
Step 3
Exam Tip
अगला कर्ण \(\sqrt{m+1}\) होगा। यदि (m+1) पूर्ण वर्ग है, तो उसका वर्गमूल पूर्ण संख्या होगा।
A. \(\sqrt{440}\) (20) और (21) के बीच है और \(\sqrt{441}=21\) है/\(\sqrt{440}\) lies between (20) and (21), and \(\sqrt{441}=21\)
Step 1
Concept
\(20^2<440<21^2\), and \(441=21^2\). Therefore \(\sqrt{441}\) is exactly at (21).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{440}\) (20) और (21) के बीच है और \(\sqrt{441}=21\) है / \(\sqrt{440}\) lies between (20) and (21), and \(\sqrt{441}=21\). \(20^2<440<21^2\), and \(441=21^2\). Therefore \(\sqrt{441}\) is exactly at (21).
Step 3
Exam Tip
\(20^2<440<21^2\) और \(441=21^2\) है। इसलिए \(\sqrt{441}\) ठीक (21) पर है।
A. दोनों (5) और (6) के बीच हैं/Both lie between (5) and (6)
Step 1
Concept
(25<27<36) and (25<32<36). Therefore both square roots 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 lie between (5) and (6). (25<27<36) and (25<32<36). Therefore both square roots lie between (5) and (6).
Step 3
Exam Tip
(25<27<36) और (25<32<36) हैं। इसलिए दोनों के वर्गमूल (5) और (6) के बीच हैं।
In Pythagoras theorem, the squares of sides are added. Therefore the new hypotenuse becomes \(\sqrt{8}\).
Step 2
Why this answer is correct
The correct answer is A. (\(\sqrt{7}\)2+12=8). In Pythagoras theorem, the squares of sides are added. Therefore the new hypotenuse becomes \(\sqrt{8}\).
Step 3
Exam Tip
पाइथागोरस प्रमेय में भुजाओं के वर्ग जुड़ते हैं। इसलिए नया कर्ण \(\sqrt{8}\) बनता है।
A. नई लंब (1) इकाई हो और समकोण बने/The new perpendicular is (1) unit and a right angle is formed
Step 1
Concept
With a (1) unit perpendicular and a right angle, (\(\sqrt{n}\)2+12=n+1) applies. This is the spiral rule.
Step 2
Why this answer is correct
The correct answer is A. नई लंब (1) इकाई हो और समकोण बने / The new perpendicular is (1) unit and a right angle is formed. With a (1) unit perpendicular and a right angle, (\(\sqrt{n}\)2+12=n+1) applies. This is the spiral rule.
Step 3
Exam Tip
(1) इकाई लंब और समकोण से (\(\sqrt{n}\)2+12=n+1) लागू होता है। यही सर्पिल का नियम है।
A. नया कर्ण \(\sqrt{169}=13\) है/The new hypotenuse is \(\sqrt{169}=13\)
Step 1
Concept
The next hypotenuse is \(\sqrt{168+1}=\sqrt{169}\). Since \(169=13^2\), its value is (13).
Step 2
Why this answer is correct
The correct answer is A. नया कर्ण \(\sqrt{169}=13\) है / The new hypotenuse is \(\sqrt{169}=13\). The next hypotenuse is \(\sqrt{168+1}=\sqrt{169}\). Since \(169=13^2\), its value is (13).
Step 3
Exam Tip
अगला कर्ण \(\sqrt{168+1}=\sqrt{169}\) होता है। \(169=13^2\), इसलिए मान (13) है।
B. दोनों (7) और (8) के बीच हैं/Both lie between (7) and (8)
Step 1
Concept
\(7^2<50<8^2\) and \(7^2<63<8^2\). Therefore both lie between (7) and (8).
Step 2
Why this answer is correct
The correct answer is B. दोनों (7) और (8) के बीच हैं / Both lie between (7) and (8). \(7^2<50<8^2\) and \(7^2<63<8^2\). Therefore both lie between (7) and (8).
Step 3
Exam Tip
\(7^2<50<8^2\) और \(7^2<63<8^2\) हैं। इसलिए दोनों (7) और (8) के बीच हैं।
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. An arc from the origin gives the correct location.
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. An arc from the origin gives the correct location.
Step 3
Exam Tip
जिस वर्गमूल को अंकित करना है, उसी कर्ण की लंबाई कंपास में ली जाती है। मूल बिंदु से चाप सही स्थान देता है।
Drawing a (1) unit perpendicular on \(\sqrt{439}\) forms \(\sqrt{440}\). The previous hypotenuse has one less number.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{439}\). Drawing a (1) unit perpendicular on \(\sqrt{439}\) forms \(\sqrt{440}\). The previous hypotenuse has one less number.
Step 3
Exam Tip
\(\sqrt{439}\) पर (1) इकाई लंब बनाने से \(\sqrt{440}\) बनता है। पिछला कर्ण एक कम संख्या का होता है।
B. \(\sqrt{150}\) (12) और (13) के बीच है और \(\sqrt{169}=13\) है/\(\sqrt{150}\) lies between (12) and (13), and \(\sqrt{169}=13\)
Step 1
Concept
\(12^2<150<13^2\), and \(169=13^2\). Therefore \(\sqrt{150}\) is less than (13).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{150}\) (12) और (13) के बीच है और \(\sqrt{169}=13\) है / \(\sqrt{150}\) lies between (12) and (13), and \(\sqrt{169}=13\). \(12^2<150<13^2\), and \(169=13^2\). Therefore \(\sqrt{150}\) is less than (13).
Step 3
Exam Tip
\(12^2<150<13^2\) और \(169=13^2\) है। इसलिए \(\sqrt{150}\) (13) से कम है।
A. क्योंकि (\(\sqrt{4}\)2+12=5)/Because (\(\sqrt{4}\)2+12=5)
Step 1
Concept
\(\sqrt{4}\) is the previous hypotenuse and (1) is the new perpendicular. Pythagoras gives hypotenuse \(\sqrt{5}\).
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (\(\sqrt{4}\)2+12=5) / Because (\(\sqrt{4}\)2+12=5). \(\sqrt{4}\) is the previous hypotenuse and (1) is the new perpendicular. Pythagoras gives hypotenuse \(\sqrt{5}\).
Step 3
Exam Tip
\(\sqrt{4}\) पिछला कर्ण है और (1) नई लंब है। पाइथागोरस से कर्ण \(\sqrt{5}\) मिलता है।
A. \(\sqrt{24}\) (4) और (5) के बीच है, \(\sqrt{26}\) (5) और (6) के बीच है/\(\sqrt{24}\) lies between (4) and (5), \(\sqrt{26}\) lies between (5) and (6)
Step 1
Concept
\(4^2<24<5^2\) and \(5^2<26<6^2\). Therefore they lie in different intervals.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{24}\) (4) और (5) के बीच है, \(\sqrt{26}\) (5) और (6) के बीच है / \(\sqrt{24}\) lies between (4) and (5), \(\sqrt{26}\) lies between (5) and (6). \(4^2<24<5^2\) and \(5^2<26<6^2\). Therefore they lie in different intervals.
Step 3
Exam Tip
\(4^2<24<5^2\) और \(5^2<26<6^2\) हैं। इसलिए दोनों अलग अंतरालों में आते हैं।
C. \(\sqrt{80}\) (8) और (9) के बीच है, \(\sqrt{82}\) (9) और (10) के बीच है क्योंकि (82>81)/\(\sqrt{80}\) lies between (8) and (9), \(\sqrt{82}\) lies between (9) and (10) because (82>81)
Step 1
Concept
Since (80<81), \(\sqrt{80}<9\), and since (81<82<100), \(\sqrt{82}\) lies between (9) and (10).
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{80}\) (8) और (9) के बीच है, \(\sqrt{82}\) (9) और (10) के बीच है क्योंकि (82>81) / \(\sqrt{80}\) lies between (8) and (9), \(\sqrt{82}\) lies between (9) and (10) because (82>81). Since (80<81), \(\sqrt{80}<9\), and since (81<82<100), \(\sqrt{82}\) lies between (9) and (10).
Step 3
Exam Tip
(80<81) होने से \(\sqrt{80}<9\), और (81<82<100) होने से \(\sqrt{82}\) (9) और (10) के बीच है।
A. (n+1) पूर्ण वर्ग नहीं है/(n+1) is not a perfect square
Step 1
Concept
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.
Step 2
Why this answer is correct
The correct answer is A. (n+1) पूर्ण वर्ग नहीं है / (n+1) is not a perfect square. 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.
Step 3
Exam Tip
पूर्णांक का वर्गमूल पूर्ण संख्या तभी होता है जब वह पूर्ण वर्ग हो। पूर्ण वर्ग न हो तो वर्गमूल अपरिमेय हो सकता है।
A. \(\sqrt{168}\) (12) और (13) के बीच, \(\sqrt{170}\) (13) और (14) के बीच है/\(\sqrt{168}\) lies between (12) and (13), \(\sqrt{170}\) lies between (13) and (14)
Step 1
Concept
Since \(168<169=13^2\), \(\sqrt{168}<13\). Since (170>169), \(\sqrt{170}>13\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{168}\) (12) और (13) के बीच, \(\sqrt{170}\) (13) और (14) के बीच है / \(\sqrt{168}\) lies between (12) and (13), \(\sqrt{170}\) lies between (13) and (14). Since \(168<169=13^2\), \(\sqrt{168}<13\). Since (170>169), \(\sqrt{170}>13\).
Step 3
Exam Tip
\(168<169=13^2\), इसलिए \(\sqrt{168}<13\)। (170>169), इसलिए \(\sqrt{170}>13\)।
A. (\(\sqrt{12}\)2+12=13), इसलिए नया कर्ण \(\sqrt{13}\)/(\(\sqrt{12}\)2+12=13), so the new hypotenuse is \(\sqrt{13}\)
Step 1
Concept
The correct reasoning is (\(\sqrt{12}\)2+12=13). Pythagoras theorem applies in the spiral.
Step 2
Why this answer is correct
The correct answer is A. (\(\sqrt{12}\)2+12=13), इसलिए नया कर्ण \(\sqrt{13}\) / (\(\sqrt{12}\)2+12=13), so the new hypotenuse is \(\sqrt{13}\). The correct reasoning is (\(\sqrt{12}\)2+12=13). Pythagoras theorem applies in the spiral.
Step 3
Exam Tip
सही तर्क (\(\sqrt{12}\)2+12=13) है। सर्पिल में पाइथागोरस प्रमेय लागू होता है।
A. अगला कर्ण \(\sqrt{36}=6\) है और \(\sqrt{37}\) (6) और (7) के बीच है/The next hypotenuse is \(\sqrt{36}=6\), and \(\sqrt{37}\) lies between (6) and (7)
Step 1
Concept
After \(\sqrt{35}\), \(\sqrt{36}=6\) is formed. Since \(6^2<37<7^2\), \(\sqrt{37}\) lies between (6) and (7).
Step 2
Why this answer is correct
The correct answer is A. अगला कर्ण \(\sqrt{36}=6\) है और \(\sqrt{37}\) (6) और (7) के बीच है / The next hypotenuse is \(\sqrt{36}=6\), and \(\sqrt{37}\) lies between (6) and (7). After \(\sqrt{35}\), \(\sqrt{36}=6\) is formed. Since \(6^2<37<7^2\), \(\sqrt{37}\) lies between (6) and (7).
Step 3
Exam Tip
\(\sqrt{35}\) के बाद \(\sqrt{36}=6\) बनता है। \(6^2<37<7^2\), इसलिए \(\sqrt{37}\) (6) और (7) के बीच है।
\(36^2=1296\) and \(37^2=1369\). The number (1368) lies between them, so \(\sqrt{1368}\) lies between (36) and (37).
Step 2
Why this answer is correct
The correct answer is B. \(36<\sqrt{1368}<37\). \(36^2=1296\) and \(37^2=1369\). The number (1368) lies between them, so \(\sqrt{1368}\) lies between (36) and (37).
Step 3
Exam Tip
\(36^2=1296\) और \(37^2=1369\) हैं। (1368) इनके बीच है, इसलिए \(\sqrt{1368}\) (36) और (37) के बीच है।
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 found using 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 found using a right triangle and Pythagoras theorem.
Step 3
Exam Tip
वर्गमूल सर्पिल में सीधे जोड़ नहीं होता। नया कर्ण समकोण त्रिभुज और पाइथागोरस से मिलता है।
A. निकटतम पूर्ण वर्गों (a-2<n<(a+1)2) की पहचान करना/Identify nearest perfect squares (a-2<n<(a+1)2)
Step 1
Concept
Nearest perfect squares give the correct interval. Then the spiral length is placed on the number line using a compass.
Step 2
Why this answer is correct
The correct answer is A. निकटतम पूर्ण वर्गों (a-2<n<(a+1)2) की पहचान करना / Identify nearest perfect squares (a-2<n<(a+1)2). Nearest perfect squares give the correct interval. Then the spiral length is placed on the number line using a compass.
Step 3
Exam Tip
निकटतम पूर्ण वर्ग सही अंतराल बताते हैं। फिर सर्पिल की लंबाई कंपास से संख्या रेखा पर रखी जाती है।
B. यह समकोण त्रिभुजों की क्रमिक रचना है जिसमें (\(\sqrt{n}\)2+12=n+1) से अगला कर्ण बनता है/It is a successive construction of right triangles where the next hypotenuse is formed by (\(\sqrt{n}\)2+12=n+1)
Step 1
Concept
A square root spiral is based on Pythagoras theorem. The previous hypotenuse and (1) unit perpendicular form the next square root.
Step 2
Why this answer is correct
The correct answer is B. यह समकोण त्रिभुजों की क्रमिक रचना है जिसमें (\(\sqrt{n}\)2+12=n+1) से अगला कर्ण बनता है / It is a successive construction of right triangles where the next hypotenuse is formed by (\(\sqrt{n}\)2+12=n+1). A square root spiral is based on Pythagoras theorem. The previous hypotenuse and (1) unit perpendicular form the next square root.
Step 3
Exam Tip
वर्गमूल सर्पिल पाइथागोरस प्रमेय पर आधारित है। पिछला कर्ण और (1) इकाई लंब मिलकर अगला वर्गमूल बनाते हैं।
