The new hypotenuse is \(\sqrt{80+1}=\sqrt{81}\). Since \(\sqrt{81}=9\), write the exact value at a perfect square.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{81}=9\). The new hypotenuse is \(\sqrt{80+1}=\sqrt{81}\). Since \(\sqrt{81}=9\), write the exact value at a perfect square.
Step 3
Exam Tip
नया कर्ण \(\sqrt{80+1}=\sqrt{81}\) होगा। \(\sqrt{81}=9\), इसलिए पूर्ण वर्ग पर सटीक मान लिखें।
B. \(\sqrt{48}\) (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
Because \(6^2<48<7^2\) and \(7^2<50<8^2\). Check nearest perfect squares while deciding intervals.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{48}\) (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). Because \(6^2<48<7^2\) and \(7^2<50<8^2\). Check nearest perfect squares while deciding intervals.
Step 3
Exam Tip
क्योंकि \(6^2<48<7^2\) और \(7^2<50<8^2\) है। अंतराल तय करते समय निकटतम पूर्ण वर्ग देखें।
C. (\sqrt{\(\sqrt{12}\)2+12}=\sqrt{13}) लिखना चाहिए/We should write (\sqrt{\(\sqrt{12}\)2+12}=\sqrt{13})
Step 1
Concept
Lengths are not added directly in a square root spiral. The correct method is to add squares using Pythagoras theorem.
Step 2
Why this answer is correct
The correct answer is C. (\sqrt{\(\sqrt{12}\)2+12}=\sqrt{13}) लिखना चाहिए / We should write (\sqrt{\(\sqrt{12}\)2+12}=\sqrt{13}). Lengths are not added directly in a square root spiral. The correct method is to add squares using Pythagoras theorem.
Step 3
Exam Tip
वर्गमूल सर्पिल में सीधे लंबाइयाँ नहीं जोड़ी जातीं। सही विधि पाइथागोरस प्रमेय से वर्गों का योग लेना है।
D. \(\sqrt{144}\), ठीक (12) पर/\(\sqrt{144}\), exactly at (12)
Step 1
Concept
The new hypotenuse is \(\sqrt{144}\). Since \(\sqrt{144}=12\), it is not in an interval but exactly at (12).
Step 2
Why this answer is correct
The correct answer is D. \(\sqrt{144}\), ठीक (12) पर / \(\sqrt{144}\), exactly at (12). The new hypotenuse is \(\sqrt{144}\). Since \(\sqrt{144}=12\), it is not in an interval but exactly at (12).
Step 3
Exam Tip
नया कर्ण \(\sqrt{144}\) है। क्योंकि \(\sqrt{144}=12\), यह किसी अंतराल में नहीं बल्कि ठीक (12) पर है।
Since (\(\sqrt{34}\)2+12=35). Therefore the previous hypotenuse for \(\sqrt{35}\) is \(\sqrt{34}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{34}\) और (1) / \(\sqrt{34}\) and (1). Since (\(\sqrt{34}\)2+12=35). Therefore the previous hypotenuse for \(\sqrt{35}\) is \(\sqrt{34}\).
Step 3
Exam Tip
(\(\sqrt{34}\)2+12=35) होता है। इसलिए \(\sqrt{35}\) के लिए पिछला कर्ण \(\sqrt{34}\) होगा।
B. \(\sqrt{99}\) (9) और (10) के बीच है/\(\sqrt{99}\) lies between (9) and (10)
Step 1
Concept
Because \(9^2<99<10^2\). Since \(\sqrt{100}=10\), \(\sqrt{99}\) is slightly less than it.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{99}\) (9) और (10) के बीच है / \(\sqrt{99}\) lies between (9) and (10). Because \(9^2<99<10^2\). Since \(\sqrt{100}=10\), \(\sqrt{99}\) is slightly less than it.
Step 3
Exam Tip
क्योंकि \(9^2<99<10^2\) है। \(\sqrt{100}=10\) है, इसलिए \(\sqrt{99}\) उससे थोड़ा कम है।
B. \(\sqrt{225}\), (15) और (16) के बीच/\(\sqrt{225}\), between (15) and (16)
Step 1
Concept
Before \(\sqrt{226}\), the previous hypotenuse was \(\sqrt{225}\). Since \(15^2<226<16^2\), the new value lies between (15) and (16).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{225}\), (15) और (16) के बीच / \(\sqrt{225}\), between (15) and (16). Before \(\sqrt{226}\), the previous hypotenuse was \(\sqrt{225}\). Since \(15^2<226<16^2\), the new value lies between (15) and (16).
Step 3
Exam Tip
\(\sqrt{226}\) से पहले \(\sqrt{225}\) था। क्योंकि \(15^2<226<16^2\), नया मान (15) और (16) के बीच है।
The new hypotenuse is not obtained directly from \(\sqrt{3}+1\). The correct basis is the sum of squares of sides.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{3}+1=\sqrt{4}\). The new hypotenuse is not obtained directly from \(\sqrt{3}+1\). The correct basis is the sum of squares of sides.
Step 3
Exam Tip
नया कर्ण सीधे \(\sqrt{3}+1\) से नहीं मिलता। सही आधार भुजाओं के वर्गों का योग है।
A. \(\sqrt{169}\) बनता है और (169) पूर्ण वर्ग है/\(\sqrt{169}\) is formed and (169) is a perfect square
Step 1
Concept
After \(\sqrt{168}\), \(\sqrt{169}\) is formed. Since \(169=13^2\), the hypotenuse value is (13).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{169}\) बनता है और (169) पूर्ण वर्ग है / \(\sqrt{169}\) is formed and (169) is a perfect square. After \(\sqrt{168}\), \(\sqrt{169}\) is formed. Since \(169=13^2\), the hypotenuse value is (13).
Step 3
Exam Tip
\(\sqrt{168}\) के बाद \(\sqrt{169}\) बनता है। \(169=13^2\) होने से कर्ण का मान (13) है।
B. \(\sqrt{242}\) (15) और (16) के बीच है, \(\sqrt{256}=16\) है/\(\sqrt{242}\) is between (15) and (16), \(\sqrt{256}=16\)
Step 1
Concept
Because \(15^2<242<16^2\) and \(256=16^2\). Therefore \(\sqrt{242}\) is less than (16).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{242}\) (15) और (16) के बीच है, \(\sqrt{256}=16\) है / \(\sqrt{242}\) is between (15) and (16), \(\sqrt{256}=16\). Because \(15^2<242<16^2\) and \(256=16^2\). Therefore \(\sqrt{242}\) is less than (16).
Step 3
Exam Tip
क्योंकि \(15^2<242<16^2\) और \(256=16^2\) है। इसलिए \(\sqrt{242}\) (16) से कम है।
C. नया कर्ण \(\sqrt{16}\) है और (4) के बराबर है/The new hypotenuse is \(\sqrt{16}\) and equals (4)
Step 1
Concept
The new hypotenuse is \(\sqrt{15+1}=\sqrt{16}\). Since \(\sqrt{16}=4\), it is a whole number.
Step 2
Why this answer is correct
The correct answer is C. नया कर्ण \(\sqrt{16}\) है और (4) के बराबर है / The new hypotenuse is \(\sqrt{16}\) and equals (4). The new hypotenuse is \(\sqrt{15+1}=\sqrt{16}\). Since \(\sqrt{16}=4\), it is a whole number.
Step 3
Exam Tip
नया कर्ण \(\sqrt{15+1}=\sqrt{16}\) होगा। \(\sqrt{16}=4\), इसलिए यह पूर्ण संख्या है।
By Pythagoras theorem, the squares of sides are added. Therefore \(\sqrt{7}\) and (1) form hypotenuse \(\sqrt{8}\).
Step 2
Why this answer is correct
The correct answer is A. (\(\sqrt{7}\)2+12=8). By Pythagoras theorem, the squares of sides are added. Therefore \(\sqrt{7}\) and (1) form hypotenuse \(\sqrt{8}\).
Step 3
Exam Tip
पाइथागोरस प्रमेय से भुजाओं के वर्ग जुड़ते हैं। इसलिए \(\sqrt{7}\) और (1) से कर्ण \(\sqrt{8}\) बनता है।
A. अगला कर्ण \(\sqrt{64}=8\) है और \(\sqrt{65}\) (8) और (9) के बीच है/The next hypotenuse is \(\sqrt{64}=8\), and \(\sqrt{65}\) is between (8) and (9)
Step 1
Concept
After \(\sqrt{63}\), \(\sqrt{64}=8\) is formed. Since \(8^2<65<9^2\), \(\sqrt{65}\) lies between (8) and (9).
Step 2
Why this answer is correct
The correct answer is A. अगला कर्ण \(\sqrt{64}=8\) है और \(\sqrt{65}\) (8) और (9) के बीच है / The next hypotenuse is \(\sqrt{64}=8\), and \(\sqrt{65}\) is between (8) and (9). After \(\sqrt{63}\), \(\sqrt{64}=8\) is formed. Since \(8^2<65<9^2\), \(\sqrt{65}\) lies between (8) and (9).
Step 3
Exam Tip
\(\sqrt{63}\) के बाद \(\sqrt{64}=8\) बनता है। \(8^2<65<9^2\), इसलिए \(\sqrt{65}\) (8) और (9) के बीच है।
The next hypotenuse is \(\sqrt{m+1}\). If (m+1) is a perfect square, its square root is a whole number.
Step 2
Why this answer is correct
The correct answer is B. हमेशा पूर्ण संख्या / Always a whole number. The next hypotenuse is \(\sqrt{m+1}\). If (m+1) is a perfect square, its square root is 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।
Because \(16^2=256\) and \(17^2=289\). The number (288) lies between them, so \(\sqrt{288}\) lies between (16) and (17).
Step 2
Why this answer is correct
The correct answer is B. \(16<\sqrt{288}<17\). Because \(16^2=256\) and \(17^2=289\). The number (288) lies between them, so \(\sqrt{288}\) lies between (16) and (17).
Step 3
Exam Tip
क्योंकि \(16^2=256\) और \(17^2=289\) हैं। (288) इनके बीच है, इसलिए \(\sqrt{288}\) (16) और (17) के बीच है।
A. क्योंकि सामान्य नियम में नई लंब (1) इकाई होती है और पिछला कर्ण \(\sqrt{47}\) होना चाहिए/Because in the usual rule the new perpendicular is (1) unit and the previous hypotenuse should be \(\sqrt{47}\)
Step 1
Concept
In the usual square root spiral, a (1) unit perpendicular is added every time. For \(\sqrt{48}\), the previous hypotenuse is \(\sqrt{47}\).
Step 2
Why this answer is correct
The correct answer is A. क्योंकि सामान्य नियम में नई लंब (1) इकाई होती है और पिछला कर्ण \(\sqrt{47}\) होना चाहिए / Because in the usual rule the new perpendicular is (1) unit and the previous hypotenuse should be \(\sqrt{47}\). In the usual square root spiral, a (1) unit perpendicular is added every time. For \(\sqrt{48}\), the previous hypotenuse is \(\sqrt{47}\).
Step 3
Exam Tip
सामान्य वर्गमूल सर्पिल में हर बार (1) इकाई लंब जोड़ी जाती है। \(\sqrt{48}\) के लिए पिछला कर्ण \(\sqrt{47}\) होता है।
B. \(\sqrt{300}\) (17) और (18) के बीच है, \(\sqrt{324}=18\) है/\(\sqrt{300}\) is between (17) and (18), \(\sqrt{324}=18\)
Step 1
Concept
Because \(17^2<300<18^2\) and \(324=18^2\). Therefore \(\sqrt{324}\) is exactly (18).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{300}\) (17) और (18) के बीच है, \(\sqrt{324}=18\) है / \(\sqrt{300}\) is between (17) and (18), \(\sqrt{324}=18\). Because \(17^2<300<18^2\) and \(324=18^2\). Therefore \(\sqrt{324}\) is exactly (18).
Step 3
Exam Tip
क्योंकि \(17^2<300<18^2\) और \(324=18^2\) है। इसलिए \(\sqrt{324}\) ठीक (18) है।
A. \(\sqrt{8}\rightarrow\sqrt{9}\rightarrow\sqrt{10}\)
Step 1
Concept
The hypotenuses increase in order in the spiral. Just before \(\sqrt{10}\) comes \(\sqrt{9}\), and before that \(\sqrt{8}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{8}\rightarrow\sqrt{9}\rightarrow\sqrt{10}\). The hypotenuses increase in order in the spiral. Just before \(\sqrt{10}\) comes \(\sqrt{9}\), and before that \(\sqrt{8}\).
Step 3
Exam Tip
सर्पिल में कर्ण क्रम से बढ़ते हैं। \(\sqrt{10}\) से ठीक पहले \(\sqrt{9}\) और उससे पहले \(\sqrt{8}\) आता है।
B. \(\sqrt{195}\) को (14) और (15) के बीच रखने की गलती/Mistakenly placing \(\sqrt{195}\) between (14) and (15)
Step 1
Concept
Actually \(13^2=169\) and \(14^2=196\). Therefore \(\sqrt{195}\) lies between (13) and (14), less than (14).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{195}\) को (14) और (15) के बीच रखने की गलती / Mistakenly placing \(\sqrt{195}\) between (14) and (15). Actually \(13^2=169\) and \(14^2=196\). Therefore \(\sqrt{195}\) lies between (13) and (14), less than (14).
Step 3
Exam Tip
वास्तव में \(13^2=169\) और \(14^2=196\) हैं। इसलिए \(\sqrt{195}\) (13) और (14) के बीच है, (14) से कम।
C. क्योंकि (\(\sqrt{n}\)2+12=n+1)/Because (\(\sqrt{n}\)2+12=n+1)
Step 1
Concept
In Pythagoras theorem, the square of the hypotenuse equals the sum of squares of the sides. This is the general rule.
Step 2
Why this answer is correct
The correct answer is C. क्योंकि (\(\sqrt{n}\)2+12=n+1) / Because (\(\sqrt{n}\)2+12=n+1). In Pythagoras theorem, the square of the hypotenuse equals the sum of squares of the sides. This is the general rule.
Step 3
Exam Tip
पाइथागोरस प्रमेय में कर्ण का वर्ग भुजाओं के वर्गों के योग के बराबर होता है। यही सामान्य नियम है।
B. \(\sqrt{50}\) (7) और (8) के बीच है, \(\sqrt{63}\) (7) और (8) के बीच है/\(\sqrt{50}\) lies between (7) and (8), \(\sqrt{63}\) lies between (7) and (8)
Step 1
Concept
Because \(7^2<50<8^2\) and \(7^2<63<8^2\). Both lie between (7) and (8).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{50}\) (7) और (8) के बीच है, \(\sqrt{63}\) (7) और (8) के बीच है / \(\sqrt{50}\) lies between (7) and (8), \(\sqrt{63}\) lies between (7) and (8). Because \(7^2<50<8^2\) and \(7^2<63<8^2\). Both lie between (7) and (8).
Step 3
Exam Tip
क्योंकि \(7^2<50<8^2\) और \(7^2<63<8^2\) है। दोनों (7) और (8) के बीच हैं।
A. वह \(\sqrt{121}=11\) है/It is \(\sqrt{121}=11\)
Step 1
Concept
The next hypotenuse is \(\sqrt{121}\). Since \(121=11^2\), the exact value of the hypotenuse is (11).
Step 2
Why this answer is correct
The correct answer is A. वह \(\sqrt{121}=11\) है / It is \(\sqrt{121}=11\). The next hypotenuse is \(\sqrt{121}\). Since \(121=11^2\), the exact value of the hypotenuse is (11).
Step 3
Exam Tip
अगला कर्ण \(\sqrt{121}\) है। \(121=11^2\), इसलिए कर्ण का सटीक मान (11) है।
B. \(\sqrt{223}\), (14) और (15) के बीच/\(\sqrt{223}\), between (14) and (15)
Step 1
Concept
\(\sqrt{224}\) is formed from \(\sqrt{223}\). Since \(14^2<224<15^2\), it lies between (14) and (15).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{223}\), (14) और (15) के बीच / \(\sqrt{223}\), between (14) and (15). \(\sqrt{224}\) is formed from \(\sqrt{223}\). Since \(14^2<224<15^2\), it lies between (14) and (15).
Step 3
Exam Tip
\(\sqrt{223}\) से \(\sqrt{224}\) बनता है। क्योंकि \(14^2<224<15^2\), यह (14) और (15) के बीच होगा।
B. कंपास में \(\sqrt{2}\) कर्ण की लंबाई लेकर मूल बिंदु से चाप खींचना/Take the \(\sqrt{2}\) hypotenuse length in compass and draw an arc from the origin
Step 1
Concept
The same length to be marked is taken in the compass. Drawing an arc from the origin gives the correct position.
Step 2
Why this answer is correct
The correct answer is B. कंपास में \(\sqrt{2}\) कर्ण की लंबाई लेकर मूल बिंदु से चाप खींचना / Take the \(\sqrt{2}\) hypotenuse length in compass and draw an arc from the origin. The same length to be marked is taken in the compass. Drawing an arc from the origin gives the correct position.
Step 3
Exam Tip
कंपास में वही लंबाई ली जाती है जिसे संख्या रेखा पर अंकित करना है। मूल बिंदु से चाप खींचना सही स्थान देता है।
B. \(\sqrt{150}\) (12) और (13) के बीच है, \(\sqrt{169}=13\) है/\(\sqrt{150}\) lies between (12) and (13), \(\sqrt{169}=13\)
Step 1
Concept
Because \(12^2<150<13^2\) 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, \(\sqrt{169}=13\) है / \(\sqrt{150}\) lies between (12) and (13), \(\sqrt{169}=13\). Because \(12^2<150<13^2\) and \(169=13^2\). Therefore \(\sqrt{150}\) is less than (13).
Step 3
Exam Tip
क्योंकि \(12^2<150<13^2\) 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 the new 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 the new hypotenuse \(\sqrt{5}\).
Step 3
Exam Tip
\(\sqrt{4}\) पिछले कर्ण की लंबाई है और (1) नई लंब है। पाइथागोरस से नया कर्ण \(\sqrt{5}\) मिलता है।
If the (k)-th hypotenuse is \(\sqrt{k}\), then \(\sqrt{25}\) is the (25)-th hypotenuse. Also, \(\sqrt{25}=5\).
Step 2
Why this answer is correct
The correct answer is B. (25)वाँ, (5) / (25)-th, (5). If the (k)-th hypotenuse is \(\sqrt{k}\), then \(\sqrt{25}\) is the (25)-th hypotenuse. Also, \(\sqrt{25}=5\).
Step 3
Exam Tip
यदि (k)वाँ कर्ण \(\sqrt{k}\) है, तो \(\sqrt{25}\) (25)वाँ कर्ण है। \(\sqrt{25}=5\) होता है।
The next hypotenuse is \(\sqrt{35+1}=\sqrt{36}\). Since \(\sqrt{36}=6\), the exact value is (6).
Step 2
Why this answer is correct
The correct answer is A. वह \(\sqrt{36}=6\) है / It is \(\sqrt{36}=6\). The next hypotenuse is \(\sqrt{35+1}=\sqrt{36}\). Since \(\sqrt{36}=6\), the exact value is (6).
Step 3
Exam Tip
अगला कर्ण \(\sqrt{35+1}=\sqrt{36}\) है। \(\sqrt{36}=6\), इसलिए सटीक मान (6) है।
B. हर नया त्रिभुज पिछले कर्ण और (1) इकाई लंब से बनने वाला समकोण त्रिभुज है/Each new triangle is a right triangle made from the previous hypotenuse and a (1) unit perpendicular
Step 1
Concept
The correct basis of a square root spiral is a right triangle and Pythagoras theorem. Directly adding lengths is wrong.
Step 2
Why this answer is correct
The correct answer is B. हर नया त्रिभुज पिछले कर्ण और (1) इकाई लंब से बनने वाला समकोण त्रिभुज है / Each new triangle is a right triangle made from the previous hypotenuse and a (1) unit perpendicular. The correct basis of a square root spiral is a right triangle and Pythagoras theorem. Directly adding lengths is wrong.
Step 3
Exam Tip
वर्गमूल सर्पिल का सही आधार समकोण त्रिभुज और पाइथागोरस प्रमेय है। सीधे लंबाइयों को जोड़ना गलत तरीका है।
B. \(\sqrt{27}\) (5) और (6) के बीच है, \(\sqrt{32}\) (5) और (6) के बीच है/\(\sqrt{27}\) lies between (5) and (6), \(\sqrt{32}\) lies between (5) and (6)
Step 1
Concept
Both \(5^2<27<6^2\) and \(5^2<32<6^2\) are true. Therefore both lie between (5) and (6).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{27}\) (5) और (6) के बीच है, \(\sqrt{32}\) (5) और (6) के बीच है / \(\sqrt{27}\) lies between (5) and (6), \(\sqrt{32}\) lies between (5) and (6). Both \(5^2<27<6^2\) and \(5^2<32<6^2\) are true. Therefore both lie between (5) and (6).
Step 3
Exam Tip
\(5^2<27<6^2\) और \(5^2<32<6^2\) दोनों सही हैं। इसलिए दोनों (5) और (6) के बीच हैं।
Because \(24^2=576\) and \(25^2=625\). The number (624) lies between them, so \(\sqrt{624}\) lies between (24) and (25).
Step 2
Why this answer is correct
The correct answer is B. \(24<\sqrt{624}<25\). Because \(24^2=576\) and \(25^2=625\). The number (624) lies between them, so \(\sqrt{624}\) lies between (24) and (25).
Step 3
Exam Tip
क्योंकि \(24^2=576\) और \(25^2=625\) हैं। (624) इनके बीच है, इसलिए \(\sqrt{624}\) (24) और (25) के बीच है।
A. (n+1) पूर्ण वर्ग नहीं है/(n+1) is not a perfect square
Step 1
Concept
The square root of a positive integer is a whole number only when it is a perfect square. If it is not a perfect square, the square root is 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 a positive integer is a whole number only when it is a perfect square. If it is not a perfect square, the square root is irrational.
Step 3
Exam Tip
किसी धनात्मक पूर्णांक का वर्गमूल पूर्ण संख्या तभी होता है जब वह पूर्ण वर्ग हो। पूर्ण वर्ग न हो तो वर्गमूल अपरिमेय होता है।
Pythagoras theorem uses the sum of squares. Therefore \(\sqrt{20}\) and (1) form hypotenuse \(\sqrt{21}\).
Step 2
Why this answer is correct
The correct answer is A. (\(\sqrt{20}\)2+12=21). Pythagoras theorem uses the sum of squares. Therefore \(\sqrt{20}\) and (1) form hypotenuse \(\sqrt{21}\).
Step 3
Exam Tip
पाइथागोरस प्रमेय में वर्गों का योग लिया जाता है। इसलिए \(\sqrt{20}\) और (1) से कर्ण \(\sqrt{21}\) बनता है।
C. \(\sqrt{168}\) (13) और (14) के बीच है, \(\sqrt{170}\) (13) और (14) के बीच है/\(\sqrt{168}\) lies between (13) and (14), \(\sqrt{170}\) lies between (13) and (14)
Step 1
Concept
Because \(13^2=169\). The number (168) is slightly less, so it is between (12) and (13), while (170) is between (13) and (14).
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{168}\) (13) और (14) के बीच है, \(\sqrt{170}\) (13) और (14) के बीच है / \(\sqrt{168}\) lies between (13) and (14), \(\sqrt{170}\) lies between (13) and (14). Because \(13^2=169\). The number (168) is slightly less, so it is between (12) and (13), while (170) is between (13) and (14).
Step 3
Exam Tip
क्योंकि \(13^2=169\) है। (168) थोड़ा कम है इसलिए (12) और (13) के बीच, जबकि (170) (13) और (14) के बीच है।
Because \(28^2=784\) and \(29^2=841\). The number (840) lies between them, so \(\sqrt{840}\) lies between (28) and (29).
Step 2
Why this answer is correct
The correct answer is B. \(28<\sqrt{840}<29\). Because \(28^2=784\) and \(29^2=841\). The number (840) lies between them, so \(\sqrt{840}\) lies between (28) and (29).
Step 3
Exam Tip
क्योंकि \(28^2=784\) और \(29^2=841\) हैं। (840) इनके बीच है, इसलिए \(\sqrt{840}\) (28) और (29) के बीच है।
A. नया कर्ण (5) है और \(\sqrt{24}\) (4) और (5) के बीच है/The new hypotenuse is (5), and \(\sqrt{24}\) lies between (4) and (5)
Step 1
Concept
After \(\sqrt{24}\), \(\sqrt{25}=5\) is formed. Also \(4^2<24<5^2\), so \(\sqrt{24}\) lies between (4) and (5).
Step 2
Why this answer is correct
The correct answer is A. नया कर्ण (5) है और \(\sqrt{24}\) (4) और (5) के बीच है / The new hypotenuse is (5), and \(\sqrt{24}\) lies between (4) and (5). After \(\sqrt{24}\), \(\sqrt{25}=5\) is formed. Also \(4^2<24<5^2\), so \(\sqrt{24}\) lies between (4) and (5).
Step 3
Exam Tip
\(\sqrt{24}\) के बाद \(\sqrt{25}=5\) बनता है। साथ ही \(4^2<24<5^2\), इसलिए \(\sqrt{24}\) (4) और (5) के बीच है।
B. 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 mark the spiral length using a compass.
Step 2
Why this answer is correct
The correct answer is B. 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 mark the spiral length using a compass.
Step 3
Exam Tip
निकटतम पूर्ण वर्गों से सही अंतराल मिलता है। इसके बाद सर्पिल की लंबाई कंपास से अंकित करें।
A. \(\sqrt{440}\) (20) और (21) के बीच है, \(\sqrt{441}=21\) है/\(\sqrt{440}\) lies between (20) and (21), \(\sqrt{441}=21\)
Step 1
Concept
Because \(20^2<440<21^2\), and \(441=21^2\). Therefore \(\sqrt{441}\) is exactly (21).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{440}\) (20) और (21) के बीच है, \(\sqrt{441}=21\) है / \(\sqrt{440}\) lies between (20) and (21), \(\sqrt{441}=21\). Because \(20^2<440<21^2\), and \(441=21^2\). Therefore \(\sqrt{441}\) is exactly (21).
Step 3
Exam Tip
क्योंकि \(20^2<440<21^2\) और \(441=21^2\) है। इसलिए \(\sqrt{441}\) ठीक (21) है।
D. नया कर्ण \(\sqrt{n}+1\) होगा/The new hypotenuse will be \(\sqrt{n}+1\)
Step 1
Concept
The new hypotenuse is not directly \(\sqrt{n}+1\). It is (\sqrt{\(\sqrt{n}\)2+12}=\sqrt{n+1}).
Step 2
Why this answer is correct
The correct answer is D. नया कर्ण \(\sqrt{n}+1\) होगा / The new hypotenuse will be \(\sqrt{n}+1\). The new hypotenuse is not directly \(\sqrt{n}+1\). It is (\sqrt{\(\sqrt{n}\)2+12}=\sqrt{n+1}).
Step 3
Exam Tip
नया कर्ण सीधे \(\sqrt{n}+1\) नहीं होता। वह (\sqrt{\(\sqrt{n}\)2+12}=\sqrt{n+1}) होता है।
Because \(30^2=900\) and \(31^2=961\). The number (960) lies between them, so \(\sqrt{960}\) lies between (30) and (31).
Step 2
Why this answer is correct
The correct answer is B. \(30<\sqrt{960}<31\). Because \(30^2=900\) and \(31^2=961\). The number (960) lies between them, so \(\sqrt{960}\) lies between (30) and (31).
Step 3
Exam Tip
क्योंकि \(30^2=900\) और \(31^2=961\) हैं। (960) इनके बीच है, इसलिए \(\sqrt{960}\) (30) और (31) के बीच 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।
