By Pythagoras the new hypotenuse is (\sqrt{\(\sqrt{13}\)2+12}=\sqrt{14}). In a square root spiral the number increases by one.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{13+1}\). By Pythagoras the new hypotenuse is (\sqrt{\(\sqrt{13}\)2+12}=\sqrt{14}). In a square root spiral the number increases by one.
Step 3
Exam Tip
पाइथागोरस से नया कर्ण (\sqrt{\(\sqrt{13}\)2+12}=\sqrt{14}) होगा। वर्गमूल सर्पिल में संख्या एक बढ़ती है।
Since (\(\sqrt{20}\)2+12=21). Therefore \(\sqrt{20}\) is the correct previous hypotenuse.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{20}\) और (1) / \(\sqrt{20}\) and (1). Since (\(\sqrt{20}\)2+12=21). Therefore \(\sqrt{20}\) is the correct previous hypotenuse.
Step 3
Exam Tip
(\(\sqrt{20}\)2+12=21) होता है। इसलिए \(\sqrt{20}\) पिछले कर्ण के रूप में सही है।
The next hypotenuse is \(\sqrt{36}\) and \(\sqrt{36}=6\). It lies exactly at (6), not between (5) and (6).
Step 2
Why this answer is correct
The correct answer is B. (6) और (7) के बीच / Between (6) and (7). The next hypotenuse is \(\sqrt{36}\) and \(\sqrt{36}=6\). It lies exactly at (6), not between (5) and (6).
Step 3
Exam Tip
अगला कर्ण \(\sqrt{36}\) है और \(\sqrt{36}=6\) होता है। यह (6) पर स्थित है, इसलिए (5) और (6) के बीच नहीं बल्कि ठीक (6) पर है।
In a square root spiral lengths are not added directly. The hypotenuse is formed by Pythagoras theorem.
Step 2
Why this answer is correct
The correct answer is A. (\sqrt{\(\sqrt{7}\)2+12}=\sqrt{8}). In a square root spiral lengths are not added directly. The hypotenuse is formed by Pythagoras theorem.
Step 3
Exam Tip
वर्गमूल सर्पिल में सीधे लंबाइयाँ नहीं जोड़ी जातीं। कर्ण पाइथागोरस प्रमेय से बनता है।
B. \(\sqrt{48}\) अपरिमेय है और \(\sqrt{49}=7\) है/\(\sqrt{48}\) is irrational and \(\sqrt{49}=7\)
Step 1
Concept
(48) is not a perfect square but (49) is a perfect square. Therefore \(\sqrt{48}\) is irrational and \(\sqrt{49}=7\).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{48}\) अपरिमेय है और \(\sqrt{49}=7\) है / \(\sqrt{48}\) is irrational and \(\sqrt{49}=7\). (48) is not a perfect square but (49) is a perfect square. Therefore \(\sqrt{48}\) is irrational and \(\sqrt{49}=7\).
Step 3
Exam Tip
(48) पूर्ण वर्ग नहीं है पर (49) पूर्ण वर्ग है। इसलिए \(\sqrt{48}\) अपरिमेय और \(\sqrt{49}=7\) है।
B. \(\sqrt{43}\) पर (1) इकाई लंब बनाना/Draw a (1) unit perpendicular on \(\sqrt{43}\)
Step 1
Concept
With \(\sqrt{43}\) and a (1) unit perpendicular, the new hypotenuse becomes \(\sqrt{44}\). The previous hypotenuse has one less number.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{43}\) पर (1) इकाई लंब बनाना / Draw a (1) unit perpendicular on \(\sqrt{43}\). With \(\sqrt{43}\) and a (1) unit perpendicular, the new hypotenuse becomes \(\sqrt{44}\). The previous hypotenuse has one less number.
Step 3
Exam Tip
\(\sqrt{43}\) के साथ (1) इकाई लंब से नया कर्ण \(\sqrt{44}\) बनता है। पिछले कर्ण की संख्या एक कम होती है।
If the new hypotenuse is \(\sqrt{n+1}\), the previous one is \(\sqrt{n}\). Therefore before \(\sqrt{82}\), it was \(\sqrt{81}\).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{81}\). If the new hypotenuse is \(\sqrt{n+1}\), the previous one is \(\sqrt{n}\). Therefore before \(\sqrt{82}\), it was \(\sqrt{81}\).
Step 3
Exam Tip
नया कर्ण \(\sqrt{n+1}\) होता है तो पिछला कर्ण \(\sqrt{n}\) होता है। इसलिए \(\sqrt{82}\) से पहले \(\sqrt{81}\) था।
A. \(\sqrt{81}=9\) और नया कर्ण \(\sqrt{82}\) है/\(\sqrt{81}=9\) and the new hypotenuse is \(\sqrt{82}\)
Step 1
Concept
\(\sqrt{81}=9\), and after adding a (1) unit perpendicular the next hypotenuse is \(\sqrt{82}\). The number increases by one.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{81}=9\) और नया कर्ण \(\sqrt{82}\) है / \(\sqrt{81}=9\) and the new hypotenuse is \(\sqrt{82}\). \(\sqrt{81}=9\), and after adding a (1) unit perpendicular the next hypotenuse is \(\sqrt{82}\). The number increases by one.
Step 3
Exam Tip
\(\sqrt{81}=9\) है और (1) इकाई लंब जोड़ने पर अगला कर्ण \(\sqrt{82}\) बनता है। क्रम में संख्या एक बढ़ती है।
A. क्योंकि \(1^2\) की जगह \(3^2\) जुड़ेगा/Because \(3^2\) will be added instead of \(1^2\)
Step 1
Concept
In the usual sequence \(1^2\) is added each time. Taking (3) units adds (9), so the sequence changes.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि \(1^2\) की जगह \(3^2\) जुड़ेगा / Because \(3^2\) will be added instead of \(1^2\). In the usual sequence \(1^2\) is added each time. Taking (3) units adds (9), so the sequence changes.
Step 3
Exam Tip
सामान्य क्रम में हर बार \(1^2\) जुड़ता है। (3) इकाई लेने पर (9) जुड़ेगा और क्रम बदल जाएगा।
B. (3) और (4) क्योंकि \(3^2<11<4^2\)/(3) and (4) because \(3^2<11<4^2\)
Step 1
Concept
Since (9<11<16). Therefore \(3<\sqrt{11}<4\) is correct.
Step 2
Why this answer is correct
The correct answer is B. (3) और (4) क्योंकि \(3^2<11<4^2\) / (3) and (4) because \(3^2<11<4^2\). Since (9<11<16). Therefore \(3<\sqrt{11}<4\) is correct.
Step 3
Exam Tip
(9<11<16) होता है। इसलिए \(3<\sqrt{11}<4\) सही है।
A. \(\sqrt{2}\) के सिरे पर (1) इकाई लंब खींचना/Draw a (1) unit perpendicular at the end of \(\sqrt{2}\)
Step 1
Concept
\(\sqrt{2}\) becomes the previous hypotenuse side. A (1) unit perpendicular at its end gives the new hypotenuse \(\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}\) के सिरे पर (1) इकाई लंब खींचना / Draw a (1) unit perpendicular at the end of \(\sqrt{2}\). \(\sqrt{2}\) becomes the previous hypotenuse side. A (1) unit perpendicular at its end gives the new hypotenuse \(\sqrt{3}\).
Step 3
Exam Tip
\(\sqrt{2}\) पिछले कर्ण की भुजा बनती है। उसके सिरे पर (1) इकाई लंब से नया कर्ण \(\sqrt{3}\) मिलता है।
\(\sqrt{143}\) is formed from \(\sqrt{142}\) and a (1) unit perpendicular. The number increases by one in the next hypotenuse.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{142}\). \(\sqrt{143}\) is formed from \(\sqrt{142}\) and a (1) unit perpendicular. The number increases by one in the next hypotenuse.
Step 3
Exam Tip
\(\sqrt{142}\) और (1) इकाई लंब से \(\sqrt{143}\) बनता है। अगले कर्ण में संख्या एक बढ़ती है।
B. \(\sqrt{39}\) (6) और (7) के बीच है/\(\sqrt{39}\) lies between (6) and (7)
Step 1
Concept
Because \(6^2<39<7^2\). Therefore \(6<\sqrt{39}<7\).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{39}\) (6) और (7) के बीच है / \(\sqrt{39}\) lies between (6) and (7). Because \(6^2<39<7^2\). Therefore \(6<\sqrt{39}<7\).
Step 3
Exam Tip
क्योंकि \(6^2<39<7^2\) है। इसलिए \(6<\sqrt{39}<7\) होगा।
\(\sqrt{7}\) is the previous hypotenuse and the new perpendicular is (1) unit. Therefore the new hypotenuse is \(\sqrt{8}\).
Step 2
Why this answer is correct
The correct answer is A. (\(\sqrt{7}\)2+12=8). \(\sqrt{7}\) is the previous hypotenuse and the new perpendicular is (1) unit. Therefore the new hypotenuse is \(\sqrt{8}\).
Step 3
Exam Tip
\(\sqrt{7}\) पिछले कर्ण की लंबाई है और नई लंब (1) इकाई है। इसलिए नया कर्ण \(\sqrt{8}\) है।
A. \(\sqrt{169}=13\) और \(\sqrt{170}\) अपरिमेय है/\(\sqrt{169}=13\) and \(\sqrt{170}\) is irrational
Step 1
Concept
(169) is a perfect square and (170) is not. Therefore \(\sqrt{169}=13\) and \(\sqrt{170}\) is irrational.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{169}=13\) और \(\sqrt{170}\) अपरिमेय है / \(\sqrt{169}=13\) and \(\sqrt{170}\) is irrational. (169) is a perfect square and (170) is not. Therefore \(\sqrt{169}=13\) and \(\sqrt{170}\) is irrational.
Step 3
Exam Tip
(169) पूर्ण वर्ग है और (170) पूर्ण वर्ग नहीं है। इसलिए \(\sqrt{169}=13\) और \(\sqrt{170}\) अपरिमेय है।
A. कंपास में \(\sqrt{n}\) कर्ण की लंबाई लेकर मूल बिंदु से चाप खींचना/Take the \(\sqrt{n}\) 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 A. कंपास में \(\sqrt{n}\) कर्ण की लंबाई लेकर मूल बिंदु से चाप खींचना / Take the \(\sqrt{n}\) 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
जिस लंबाई को अंकित करना है, वही कंपास में ली जाती है। मूल बिंदु से चाप खींचने पर सही स्थान मिलता है।
A. \(\sqrt{56}\), (7) और (8) के बीच/\(\sqrt{56}\), between (7) and (8)
Step 1
Concept
The next hypotenuse is \(\sqrt{56}\). Since \(7^2<56<8^2\), it lies between (7) and (8).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{56}\), (7) और (8) के बीच / \(\sqrt{56}\), between (7) and (8). The next hypotenuse is \(\sqrt{56}\). Since \(7^2<56<8^2\), it lies between (7) and (8).
Step 3
Exam Tip
अगला कर्ण \(\sqrt{56}\) है। क्योंकि \(7^2<56<8^2\), यह (7) और (8) के बीच होगा।
A. सेट स्क्वायर, \(90^\circ\) कोण बनाने के लिए/Set square, to make a \(90^\circ\) angle
Step 1
Concept
A set square can be used to make a right angle. If the right angle is correct, Pythagoras gives the correct hypotenuse.
Step 2
Why this answer is correct
The correct answer is A. सेट स्क्वायर, \(90^\circ\) कोण बनाने के लिए / Set square, to make a \(90^\circ\) angle. A set square can be used to make a right angle. If the right angle is correct, Pythagoras gives the correct hypotenuse.
Step 3
Exam Tip
सेट स्क्वायर से समकोण बनाया जा सकता है। समकोण सही होगा तो पाइथागोरस से कर्ण सही मिलेगा।
B. \(\sqrt{2}\) (1) और (2) के बीच, \(\sqrt{5}\) (2) और (3) के बीच है/\(\sqrt{2}\) is between (1) and (2), \(\sqrt{5}\) is between (2) and (3)
Step 1
Concept
\(1^2<2<2^2\) and \(2^2<5<3^2\). Therefore they lie in different intervals.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{2}\) (1) और (2) के बीच, \(\sqrt{5}\) (2) और (3) के बीच है / \(\sqrt{2}\) is between (1) and (2), \(\sqrt{5}\) is between (2) and (3). \(1^2<2<2^2\) and \(2^2<5<3^2\). Therefore they lie in different intervals.
Step 3
Exam Tip
\(1^2<2<2^2\) और \(2^2<5<3^2\) है। इसलिए दोनों की स्थिति अलग अंतरालों में है।
A whole number square root is obtained only when (n) is a perfect square. For example \(\sqrt{64}=8\).
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 \(\sqrt{64}=8\).
Step 3
Exam Tip
पूर्ण संख्या वर्गमूल तभी मिलता है जब (n) पूर्ण वर्ग हो। जैसे \(\sqrt{64}=8\)।
A. नए कर्णों की लंबाइयाँ गलत हो सकती हैं/The lengths of new hypotenuses may become incorrect
Step 1
Concept
The (1) unit perpendicular must be correct in construction. Wrong measurement will not make the next square root correctly.
Step 2
Why this answer is correct
The correct answer is A. नए कर्णों की लंबाइयाँ गलत हो सकती हैं / The lengths of new hypotenuses may become incorrect. The (1) unit perpendicular must be correct in construction. Wrong measurement will not make the next square root correctly.
Step 3
Exam Tip
निर्माण में (1) इकाई लंब सही होना जरूरी है। गलत माप से अगला वर्गमूल सही नहीं बनेगा।
B. अपरिमेय संख्या और (5) तथा (6) के बीच/Irrational number and between (5) and (6)
Step 1
Concept
(28) is not a perfect square and \(5^2<28<6^2\). Therefore \(\sqrt{28}\) is irrational and lies between (5) and (6).
Step 2
Why this answer is correct
The correct answer is B. अपरिमेय संख्या और (5) तथा (6) के बीच / Irrational number and between (5) and (6). (28) is not a perfect square and \(5^2<28<6^2\). Therefore \(\sqrt{28}\) is irrational and lies between (5) and (6).
Step 3
Exam Tip
(28) पूर्ण वर्ग नहीं है और \(5^2<28<6^2\) है। इसलिए \(\sqrt{28}\) अपरिमेय है और (5) तथा (6) के बीच है।
In a right triangle the square of the hypotenuse equals the sum of squares of the sides. Therefore the hypotenuse is \(\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. क्योंकि \(1^2+1^2=2\) / Because \(1^2+1^2=2\). In a right triangle the square of the hypotenuse equals the sum of squares of the sides. Therefore the hypotenuse is \(\sqrt{2}\).
Step 3
Exam Tip
समकोण त्रिभुज में कर्ण का वर्ग भुजाओं के वर्गों के योग के बराबर होता है। इसलिए कर्ण \(\sqrt{2}\) है।
A. \(\sqrt{100}=10\) और \(\sqrt{101}\) (10) तथा (11) के बीच है/\(\sqrt{100}=10\) and \(\sqrt{101}\) is between (10) and (11)
Step 1
Concept
\(\sqrt{100}=10\) and \(10^2<101<11^2\). Therefore \(\sqrt{101}\) lies between (10) and (11).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{100}=10\) और \(\sqrt{101}\) (10) तथा (11) के बीच है / \(\sqrt{100}=10\) and \(\sqrt{101}\) is between (10) and (11). \(\sqrt{100}=10\) and \(10^2<101<11^2\). Therefore \(\sqrt{101}\) lies between (10) and (11).
Step 3
Exam Tip
\(\sqrt{100}=10\) है और \(10^2<101<11^2\) है। इसलिए \(\sqrt{101}\) (10) और (11) के बीच है।
A. \(\sqrt{58}\), (7) और (8) के बीच/\(\sqrt{58}\), between (7) and (8)
Step 1
Concept
\(\sqrt{59}\) is formed from \(\sqrt{58}\), and \(7^2<59<8^2\). Therefore it lies between (7) and (8).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{58}\), (7) और (8) के बीच / \(\sqrt{58}\), between (7) and (8). \(\sqrt{59}\) is formed from \(\sqrt{58}\), and \(7^2<59<8^2\). Therefore it lies between (7) and (8).
Step 3
Exam Tip
\(\sqrt{58}\) से \(\sqrt{59}\) बनता है और \(7^2<59<8^2\) है। इसलिए यह (7) और (8) के बीच है।
A. \(\sqrt{1}\rightarrow\sqrt{2}\rightarrow\sqrt{3}\rightarrow\sqrt{4}\)
Step 1
Concept
In a square root spiral the hypotenuses are formed in order. Skipping a step is not correct in the usual construction.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{1}\rightarrow\sqrt{2}\rightarrow\sqrt{3}\rightarrow\sqrt{4}\). In a square root spiral the hypotenuses are formed in order. Skipping a step is not correct in the usual construction.
Step 3
Exam Tip
वर्गमूल सर्पिल में कर्ण क्रम से बनते हैं। कोई चरण छोड़ना सामान्य निर्माण में सही नहीं है।
(\(\sqrt{5}\)2+12=6). Therefore sides \(\sqrt{5}\) and (1) form hypotenuse \(\sqrt{6}\).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{5}\) और (1) / \(\sqrt{5}\) and (1). (\(\sqrt{5}\)2+12=6). Therefore sides \(\sqrt{5}\) and (1) form hypotenuse \(\sqrt{6}\).
Step 3
Exam Tip
(\(\sqrt{5}\)2+12=6) होता है। इसलिए \(\sqrt{5}\) और (1) से कर्ण \(\sqrt{6}\) बनेगा।
B. अपरिमेय संख्या, (3) और (4) के बीच/Irrational number, between (3) and (4)
Step 1
Concept
(12) is not a perfect square and \(3^2<12<4^2\). Therefore it is irrational and lies between (3) and (4).
Step 2
Why this answer is correct
The correct answer is B. अपरिमेय संख्या, (3) और (4) के बीच / Irrational number, between (3) and (4). (12) is not a perfect square and \(3^2<12<4^2\). Therefore it is irrational and lies between (3) and (4).
Step 3
Exam Tip
(12) पूर्ण वर्ग नहीं है और \(3^2<12<4^2\) है। इसलिए यह अपरिमेय है और (3) तथा (4) के बीच है।
A. पिछले कर्ण पर (1) इकाई लंब जोड़ना/Adding a (1) unit perpendicular to the previous hypotenuse
Step 1
Concept
At each step the previous hypotenuse becomes one side and (1) unit perpendicular becomes the other side. This gives the next hypotenuse.
Step 2
Why this answer is correct
The correct answer is A. पिछले कर्ण पर (1) इकाई लंब जोड़ना / Adding a (1) unit perpendicular to the previous hypotenuse. At each step the previous hypotenuse becomes one side and (1) unit perpendicular becomes the other side. This gives the next hypotenuse.
Step 3
Exam Tip
हर चरण में पिछला कर्ण एक भुजा बनता है और (1) इकाई लंब दूसरी भुजा। इससे अगला कर्ण बनता है।
D. कर्ण को सीधे पिछले कर्ण में (1) जोड़कर बनाना/Making the hypotenuse by directly adding (1) to the previous hypotenuse
Step 1
Concept
The hypotenuse is not made by direct addition. It is found using a right triangle and Pythagoras theorem.
Step 2
Why this answer is correct
The correct answer is D. कर्ण को सीधे पिछले कर्ण में (1) जोड़कर बनाना / Making the hypotenuse by directly adding (1) to the previous hypotenuse. The hypotenuse is not made by direct addition. It is found using a right triangle and Pythagoras theorem.
Step 3
Exam Tip
कर्ण सीधे जोड़ से नहीं बनता। उसे समकोण त्रिभुज और पाइथागोरस प्रमेय से निकाला जाता है।
A. \(\sqrt{225}=15\) और अगला कर्ण \(\sqrt{226}\) है/\(\sqrt{225}=15\) and the next hypotenuse is \(\sqrt{226}\)
Step 1
Concept
\(\sqrt{225}=15\). In the next step, adding a (1) unit perpendicular forms \(\sqrt{226}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{225}=15\) और अगला कर्ण \(\sqrt{226}\) है / \(\sqrt{225}=15\) and the next hypotenuse is \(\sqrt{226}\). \(\sqrt{225}=15\). In the next step, adding a (1) unit perpendicular forms \(\sqrt{226}\).
Step 3
Exam Tip
\(\sqrt{225}=15\) होता है। अगले चरण में (1) इकाई लंब जोड़ने से \(\sqrt{226}\) बनता है।
A. ताकि हर चरण में (\(\sqrt{n}\)2+12=n+1) लागू हो सके/So that (\(\sqrt{n}\)2+12=n+1) can apply at each step
Step 1
Concept
With a (1) unit perpendicular and a \(90^\circ\) angle, Pythagoras theorem applies correctly. This forms successive square roots.
Step 2
Why this answer is correct
The correct answer is A. ताकि हर चरण में (\(\sqrt{n}\)2+12=n+1) लागू हो सके / So that (\(\sqrt{n}\)2+12=n+1) can apply at each step. With a (1) unit perpendicular and a \(90^\circ\) angle, Pythagoras theorem applies correctly. This forms successive square roots.
Step 3
Exam Tip
(1) इकाई लंब और \(90^\circ\) कोण से पाइथागोरस प्रमेय सही लागू होता है। इससे क्रमिक वर्गमूल बनते हैं।
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 हैं / Both are irrational. 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।
A. पिछला कर्ण \(\sqrt{n}\) और नई लंब (1) है/Previous hypotenuse is \(\sqrt{n}\) and new perpendicular is (1)
Step 1
Concept
(\(\sqrt{n}\)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) है / Previous hypotenuse is \(\sqrt{n}\) and new perpendicular is (1). (\(\sqrt{n}\)2+12=n+1). Therefore the new hypotenuse becomes \(\sqrt{n+1}\).
Step 3
Exam Tip
(\(\sqrt{n}\)2+12=n+1) होता है। इसलिए नया कर्ण \(\sqrt{n+1}\) बनता है।
A. \(\sqrt{72}\) (8) और (9) के बीच है और \(\sqrt{81}=9\)/\(\sqrt{72}\) is between (8) and (9) and \(\sqrt{81}=9\)
Step 1
Concept
\(8^2<72<9^2\) and \(81=9^2\). Therefore the statement is correct.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{72}\) (8) और (9) के बीच है और \(\sqrt{81}=9\) / \(\sqrt{72}\) is between (8) and (9) and \(\sqrt{81}=9\). \(8^2<72<9^2\) and \(81=9^2\). Therefore the statement is correct.
Step 3
Exam Tip
\(8^2<72<9^2\) और \(81=9^2\) है। इसलिए कथन सही है।
Adding a (1) unit perpendicular to \(\sqrt{254}\) gives \(\sqrt{255}\). The previous hypotenuse has one less number.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{254}\). Adding a (1) unit perpendicular to \(\sqrt{254}\) gives \(\sqrt{255}\). The previous hypotenuse has one less number.
Step 3
Exam Tip
\(\sqrt{254}\) में (1) इकाई लंब जोड़ने पर \(\sqrt{255}\) मिलता है। पिछले कर्ण की संख्या एक कम होती है।
A. समकोण त्रिभुजों की क्रमिक श्रृंखला जिसमें पिछला कर्ण और (1) इकाई लंब अगला वर्गमूल बनाते हैं/A successive chain of right triangles where previous hypotenuse and (1) unit perpendicular form the next square root
Step 1
Concept
The rule of the square root spiral is based on Pythagoras theorem and a (1) unit perpendicular. This forms successive square roots.
Step 2
Why this answer is correct
The correct answer is A. समकोण त्रिभुजों की क्रमिक श्रृंखला जिसमें पिछला कर्ण और (1) इकाई लंब अगला वर्गमूल बनाते हैं / A successive chain of right triangles where previous hypotenuse and (1) unit perpendicular form the next square root. The rule of the square root spiral is based on Pythagoras theorem and a (1) unit perpendicular. This forms successive square roots.
Step 3
Exam Tip
वर्गमूल सर्पिल का नियम पाइथागोरस प्रमेय और (1) इकाई लंब पर आधारित है। इसी से क्रमिक वर्गमूल बनते हैं।
A. \(\sqrt{121}\), (11) पर/\(\sqrt{121}\), at (11)
Step 1
Concept
The new hypotenuse is \(\sqrt{120+1}=\sqrt{121}\), and \(\sqrt{121}=11\). When a perfect square appears, write its exact value.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{121}\), (11) पर / \(\sqrt{121}\), at (11). The new hypotenuse is \(\sqrt{120+1}=\sqrt{121}\), and \(\sqrt{121}=11\). When a perfect square appears, write its exact value.
Step 3
Exam Tip
नया कर्ण \(\sqrt{120+1}=\sqrt{121}\) होगा और \(\sqrt{121}=11\) है। पूर्ण वर्ग दिखे तो उसका सटीक मान लिखें।
