Lengths are not added directly in a square root spiral. The correct method is to add squares by Pythagoras theorem.
Step 2
Why this answer is correct
The correct answer is B. (\sqrt{\(\sqrt{52}\)2+12}=\sqrt{53}). Lengths are not added directly in a square root spiral. The correct method is to add squares by Pythagoras theorem.
Step 3
Exam Tip
वर्गमूल सर्पिल में सीधे लंबाइयाँ नहीं जोड़ी जातीं। सही तरीका पाइथागोरस प्रमेय से वर्गों का योग लेना है।
A. \(\sqrt{625}\), ठीक (25) पर/\(\sqrt{625}\), exactly at (25)
Step 1
Concept
The new hypotenuse is \(\sqrt{625}\). Since \(\sqrt{625}=25\), it will be located exactly at (25).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{625}\), ठीक (25) पर / \(\sqrt{625}\), exactly at (25). The new hypotenuse is \(\sqrt{625}\). Since \(\sqrt{625}=25\), it will be located exactly at (25).
Step 3
Exam Tip
नया कर्ण \(\sqrt{625}\) है। क्योंकि \(\sqrt{625}=25\), यह ठीक (25) पर स्थित होगा।
B. \(\sqrt{120}\) (10) और (11) के बीच है, \(\sqrt{122}\) (11) और (12) के बीच है/\(\sqrt{120}\) lies between (10) and (11), \(\sqrt{122}\) lies between (11) and (12)
Step 1
Concept
Since \(120<121=11^2\), \(\sqrt{120}<11\). Since (122>121), \(\sqrt{122}>11\).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{120}\) (10) और (11) के बीच है, \(\sqrt{122}\) (11) और (12) के बीच है / \(\sqrt{120}\) lies between (10) and (11), \(\sqrt{122}\) lies between (11) and (12). Since \(120<121=11^2\), \(\sqrt{120}<11\). Since (122>121), \(\sqrt{122}>11\).
Step 3
Exam Tip
\(120<121=11^2\), इसलिए \(\sqrt{120}<11\)। (122>121), इसलिए \(\sqrt{122}>11\)।
If the (k)-th hypotenuse is \(\sqrt{k}\), then \(\sqrt{81}\) is the (81)-th hypotenuse. Also, \(\sqrt{81}=9\).
Step 2
Why this answer is correct
The correct answer is B. (81)वाँ, (9) / (81)-th, (9). If the (k)-th hypotenuse is \(\sqrt{k}\), then \(\sqrt{81}\) is the (81)-th hypotenuse. Also, \(\sqrt{81}=9\).
Step 3
Exam Tip
यदि (k)वाँ कर्ण \(\sqrt{k}\) है, तो \(\sqrt{81}\) (81)वाँ कर्ण है। \(\sqrt{81}=9\) होता है।
(\(\sqrt{209}\)2+12=210). So the previous hypotenuse for \(\sqrt{210}\) is \(\sqrt{209}\).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{209}\) और (1) / \(\sqrt{209}\) and (1). (\(\sqrt{209}\)2+12=210). So the previous hypotenuse for \(\sqrt{210}\) is \(\sqrt{209}\).
Step 3
Exam Tip
(\(\sqrt{209}\)2+12=210) है। इसलिए \(\sqrt{210}\) के लिए पिछला कर्ण \(\sqrt{209}\) होगा।
B. \(\sqrt{8}\rightarrow\sqrt{9}\rightarrow\sqrt{10}\rightarrow\sqrt{11}\rightarrow\sqrt{12}\)
Step 1
Concept
Hypotenuses increase one by one in the spiral. In the usual construction, intermediate steps are not skipped.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{8}\rightarrow\sqrt{9}\rightarrow\sqrt{10}\rightarrow\sqrt{11}\rightarrow\sqrt{12}\). Hypotenuses increase one by one in the spiral. In the usual construction, intermediate steps are not skipped.
Step 3
Exam Tip
सर्पिल में कर्ण क्रमिक रूप से एक-एक बढ़ते हैं। सामान्य निर्माण में बीच के चरण नहीं छोड़े जाते।
A. \(\sqrt{255}\) (15) और (16) के बीच है, \(\sqrt{257}\) (16) और (17) के बीच है/\(\sqrt{255}\) lies between (15) and (16), \(\sqrt{257}\) lies between (16) and (17)
Step 1
Concept
Since \(255<256=16^2\), \(\sqrt{255}<16\). Since (257>256), \(\sqrt{257}>16\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{255}\) (15) और (16) के बीच है, \(\sqrt{257}\) (16) और (17) के बीच है / \(\sqrt{255}\) lies between (15) and (16), \(\sqrt{257}\) lies between (16) and (17). Since \(255<256=16^2\), \(\sqrt{255}<16\). Since (257>256), \(\sqrt{257}>16\).
Step 3
Exam Tip
\(255<256=16^2\), इसलिए \(\sqrt{255}<16\)। (257>256), इसलिए \(\sqrt{257}>16\)।
A. अगला कर्ण \(\sqrt{324}=18\) है और \(\sqrt{325}\) (18) और (19) के बीच है/The next hypotenuse is \(\sqrt{324}=18\), and \(\sqrt{325}\) lies between (18) and (19)
Step 1
Concept
After \(\sqrt{323}\), \(\sqrt{324}=18\) is formed. Since \(18^2<325<19^2\), \(\sqrt{325}\) lies between (18) and (19).
Step 2
Why this answer is correct
The correct answer is A. अगला कर्ण \(\sqrt{324}=18\) है और \(\sqrt{325}\) (18) और (19) के बीच है / The next hypotenuse is \(\sqrt{324}=18\), and \(\sqrt{325}\) lies between (18) and (19). After \(\sqrt{323}\), \(\sqrt{324}=18\) is formed. Since \(18^2<325<19^2\), \(\sqrt{325}\) lies between (18) and (19).
Step 3
Exam Tip
\(\sqrt{323}\) के बाद \(\sqrt{324}=18\) बनता है। \(18^2<325<19^2\), इसलिए \(\sqrt{325}\) (18) और (19) 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. \(\sqrt{1680}\) (40) और (41) के बीच है और \(\sqrt{1681}=41\) है/\(\sqrt{1680}\) lies between (40) and (41), and \(\sqrt{1681}=41\)
Step 1
Concept
\(40^2<1680<41^2\), and \(1681=41^2\). Therefore \(\sqrt{1681}\) is exactly (41).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{1680}\) (40) और (41) के बीच है और \(\sqrt{1681}=41\) है / \(\sqrt{1680}\) lies between (40) and (41), and \(\sqrt{1681}=41\). \(40^2<1680<41^2\), and \(1681=41^2\). Therefore \(\sqrt{1681}\) is exactly (41).
Step 3
Exam Tip
\(40^2<1680<41^2\) और \(1681=41^2\) है। इसलिए \(\sqrt{1681}\) ठीक (41) है।
A. \(\sqrt{50}\) (7) और (8) के बीच है, \(\sqrt{80}\) (8) और (9) के बीच है/\(\sqrt{50}\) lies between (7) and (8), \(\sqrt{80}\) lies between (8) and (9)
Step 1
Concept
\(7^2<50<8^2\) and \(8^2<80<9^2\). Therefore they are in different intervals.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{50}\) (7) और (8) के बीच है, \(\sqrt{80}\) (8) और (9) के बीच है / \(\sqrt{50}\) lies between (7) and (8), \(\sqrt{80}\) lies between (8) and (9). \(7^2<50<8^2\) and \(8^2<80<9^2\). Therefore they are in different intervals.
Step 3
Exam Tip
\(7^2<50<8^2\) और \(8^2<80<9^2\) हैं। इसलिए दोनों अलग अंतरालों में हैं।
In Pythagoras theorem, the squares of sides are added. Therefore the new hypotenuse becomes \(\sqrt{42}\).
Step 2
Why this answer is correct
The correct answer is A. (\(\sqrt{41}\)2+12=42). In Pythagoras theorem, the squares of sides are added. Therefore the new hypotenuse becomes \(\sqrt{42}\).
Step 3
Exam Tip
पाइथागोरस प्रमेय में भुजाओं के वर्ग जुड़ते हैं। इसलिए नया कर्ण \(\sqrt{42}\) बनता है।
A. नया कर्ण \(\sqrt{289}=17\) है/The new hypotenuse is \(\sqrt{289}=17\)
Step 1
Concept
The next hypotenuse is \(\sqrt{288+1}=\sqrt{289}\). Since \(289=17^2\), its value is (17).
Step 2
Why this answer is correct
The correct answer is A. नया कर्ण \(\sqrt{289}=17\) है / The new hypotenuse is \(\sqrt{289}=17\). The next hypotenuse is \(\sqrt{288+1}=\sqrt{289}\). Since \(289=17^2\), its value is (17).
Step 3
Exam Tip
अगला कर्ण \(\sqrt{288+1}=\sqrt{289}\) होता है। \(289=17^2\), इसलिए मान (17) है।
A. \(\sqrt{399}\) (19) और (20) के बीच है, \(\sqrt{401}\) (20) और (21) के बीच है/\(\sqrt{399}\) lies between (19) and (20), \(\sqrt{401}\) lies between (20) and (21)
Step 1
Concept
Since \(399<400=20^2\), \(\sqrt{399}<20\). Since (401>400), \(\sqrt{401}>20\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{399}\) (19) और (20) के बीच है, \(\sqrt{401}\) (20) और (21) के बीच है / \(\sqrt{399}\) lies between (19) and (20), \(\sqrt{401}\) lies between (20) and (21). Since \(399<400=20^2\), \(\sqrt{399}<20\). Since (401>400), \(\sqrt{401}>20\).
Step 3
Exam Tip
\(399<400=20^2\), इसलिए \(\sqrt{399}<20\)। (401>400), इसलिए \(\sqrt{401}>20\)।
A. \(\sqrt{3}\) कर्ण की लंबाई कंपास में लेकर मूल बिंदु से चाप खींचना/Take the \(\sqrt{3}\) 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. \(\sqrt{3}\) कर्ण की लंबाई कंपास में लेकर मूल बिंदु से चाप खींचना / Take the \(\sqrt{3}\) 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 \(\sqrt{839}\) forms \(\sqrt{840}\). The previous hypotenuse has one less number.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{839}\). Drawing a (1) unit perpendicular on \(\sqrt{839}\) forms \(\sqrt{840}\). The previous hypotenuse has one less number.
Step 3
Exam Tip
\(\sqrt{839}\) पर (1) इकाई लंब बनाने से \(\sqrt{840}\) बनता है। पिछला कर्ण एक कम संख्या का होता है।
A. \(\sqrt{960}\) (30) और (31) के बीच है और \(\sqrt{1024}=32\) है/\(\sqrt{960}\) lies between (30) and (31), and \(\sqrt{1024}=32\)
Step 1
Concept
\(30^2<960<31^2\), and \(1024=32^2\). Therefore the first statement is correct.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{960}\) (30) और (31) के बीच है और \(\sqrt{1024}=32\) है / \(\sqrt{960}\) lies between (30) and (31), and \(\sqrt{1024}=32\). \(30^2<960<31^2\), and \(1024=32^2\). Therefore the first statement is correct.
Step 3
Exam Tip
\(30^2<960<31^2\) और \(1024=32^2\) है। इसलिए तुलना में पहला कथन सही है।
A. क्योंकि (\(\sqrt{9}\)2+12=10)/Because (\(\sqrt{9}\)2+12=10)
Step 1
Concept
\(\sqrt{9}\) is the previous hypotenuse and (1) is the new perpendicular. Pythagoras gives hypotenuse \(\sqrt{10}\).
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (\(\sqrt{9}\)2+12=10) / Because (\(\sqrt{9}\)2+12=10). \(\sqrt{9}\) is the previous hypotenuse and (1) is the new perpendicular. Pythagoras gives hypotenuse \(\sqrt{10}\).
Step 3
Exam Tip
\(\sqrt{9}\) पिछला कर्ण है और (1) नई लंब है। पाइथागोरस से कर्ण \(\sqrt{10}\) मिलता है।
A. \(\sqrt{35}\) (5) और (6) के बीच है, \(\sqrt{37}\) (6) और (7) के बीच है/\(\sqrt{35}\) lies between (5) and (6), \(\sqrt{37}\) lies between (6) and (7)
Step 1
Concept
Since \(35<36=6^2\), \(\sqrt{35}<6\). Since (37>36), \(\sqrt{37}>6\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{35}\) (5) और (6) के बीच है, \(\sqrt{37}\) (6) और (7) के बीच है / \(\sqrt{35}\) lies between (5) and (6), \(\sqrt{37}\) lies between (6) and (7). Since \(35<36=6^2\), \(\sqrt{35}<6\). Since (37>36), \(\sqrt{37}>6\).
Step 3
Exam Tip
\(35<36=6^2\), इसलिए \(\sqrt{35}<6\)। (37>36), इसलिए \(\sqrt{37}>6\)।
A. \(\sqrt{48}\) (6) और (7) के बीच है, \(\sqrt{50}\) (7) और (8) के बीच है/\(\sqrt{48}\) lies between (6) and (7), \(\sqrt{50}\) lies between (7) and (8)
Step 1
Concept
Since \(48<49=7^2\), \(\sqrt{48}<7\). Since (50>49), \(\sqrt{50}>7\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{48}\) (6) और (7) के बीच है, \(\sqrt{50}\) (7) और (8) के बीच है / \(\sqrt{48}\) lies between (6) and (7), \(\sqrt{50}\) lies between (7) and (8). Since \(48<49=7^2\), \(\sqrt{48}<7\). Since (50>49), \(\sqrt{50}>7\).
Step 3
Exam Tip
\(48<49=7^2\), इसलिए \(\sqrt{48}<7\)। (50>49), इसलिए \(\sqrt{50}>7\)।
A. \(\sqrt{9999}\) (99) और (100) के बीच है, \(\sqrt{10000}=100\) है/\(\sqrt{9999}\) lies between (99) and (100), and \(\sqrt{10000}=100\)
Step 1
Concept
\(99^2<9999<100^2\) and \(10000=100^2\). Therefore \(\sqrt{9999}\) is slightly less than (100).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{9999}\) (99) और (100) के बीच है, \(\sqrt{10000}=100\) है / \(\sqrt{9999}\) lies between (99) and (100), and \(\sqrt{10000}=100\). \(99^2<9999<100^2\) and \(10000=100^2\). Therefore \(\sqrt{9999}\) is slightly less than (100).
Step 3
Exam Tip
\(99^2<9999<100^2\) और \(10000=100^2\) है। इसलिए \(\sqrt{9999}\) (100) से थोड़ा कम है।
A. (\(\sqrt{72}\)2+12=73), इसलिए नया कर्ण \(\sqrt{73}\)/(\(\sqrt{72}\)2+12=73), so the new hypotenuse is \(\sqrt{73}\)
Step 1
Concept
The correct reasoning is (\(\sqrt{72}\)2+12=73). Pythagoras theorem applies in the spiral.
Step 2
Why this answer is correct
The correct answer is A. (\(\sqrt{72}\)2+12=73), इसलिए नया कर्ण \(\sqrt{73}\) / (\(\sqrt{72}\)2+12=73), so the new hypotenuse is \(\sqrt{73}\). The correct reasoning is (\(\sqrt{72}\)2+12=73). Pythagoras theorem applies in the spiral.
Step 3
Exam Tip
सही तर्क (\(\sqrt{72}\)2+12=73) है। सर्पिल में पाइथागोरस प्रमेय लागू होता है।
A. अगला कर्ण \(\sqrt{16}=4\) है और \(\sqrt{17}\) (4) और (5) के बीच है/The next hypotenuse is \(\sqrt{16}=4\), and \(\sqrt{17}\) lies between (4) and (5)
Step 1
Concept
After \(\sqrt{15}\), \(\sqrt{16}=4\) is formed. Since \(4^2<17<5^2\), \(\sqrt{17}\) lies between (4) and (5).
Step 2
Why this answer is correct
The correct answer is A. अगला कर्ण \(\sqrt{16}=4\) है और \(\sqrt{17}\) (4) और (5) के बीच है / The next hypotenuse is \(\sqrt{16}=4\), and \(\sqrt{17}\) lies between (4) and (5). After \(\sqrt{15}\), \(\sqrt{16}=4\) is formed. Since \(4^2<17<5^2\), \(\sqrt{17}\) lies between (4) and (5).
Step 3
Exam Tip
\(\sqrt{15}\) के बाद \(\sqrt{16}=4\) बनता है। \(4^2<17<5^2\), इसलिए \(\sqrt{17}\) (4) और (5) के बीच है।
\(58^2=3364\) and \(59^2=3481\). The number (3480) lies between them, so \(\sqrt{3480}\) lies between (58) and (59).
Step 2
Why this answer is correct
The correct answer is B. \(58<\sqrt{3480}<59\). \(58^2=3364\) and \(59^2=3481\). The number (3480) lies between them, so \(\sqrt{3480}\) lies between (58) and (59).
Step 3
Exam Tip
\(58^2=3364\) और \(59^2=3481\) हैं। (3480) इनके बीच है, इसलिए \(\sqrt{3480}\) (58) और (59) के बीच है।
A. \(\sqrt{2207}\) (46) और (47) के बीच है और \(\sqrt{2209}=47\) है/\(\sqrt{2207}\) lies between (46) and (47), and \(\sqrt{2209}=47\)
Step 1
Concept
\(46^2<2207<47^2\), and \(2209=47^2\). Therefore \(\sqrt{2209}\) is exactly at (47).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2207}\) (46) और (47) के बीच है और \(\sqrt{2209}=47\) है / \(\sqrt{2207}\) lies between (46) and (47), and \(\sqrt{2209}=47\). \(46^2<2207<47^2\), and \(2209=47^2\). Therefore \(\sqrt{2209}\) is exactly at (47).
Step 3
Exam Tip
\(46^2<2207<47^2\) और \(2209=47^2\) है। इसलिए \(\sqrt{2209}\) ठीक (47) पर है।
If the new hypotenuse is \(\sqrt{n+1}\), the previous hypotenuse is \(\sqrt{n}\). Therefore before \(\sqrt{4096}\), it was \(\sqrt{4095}\).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{4095}\). If the new hypotenuse is \(\sqrt{n+1}\), the previous hypotenuse is \(\sqrt{n}\). Therefore before \(\sqrt{4096}\), it was \(\sqrt{4095}\).
Step 3
Exam Tip
नया कर्ण \(\sqrt{n+1}\) हो तो पिछला कर्ण \(\sqrt{n}\) होता है। इसलिए \(\sqrt{4096}\) से पहले \(\sqrt{4095}\) था।
