Here (OC-2=\(\sqrt{2}\)2+12=3), so \(OC=\sqrt{3}\). At each new step, (1) is added to the previous square under the root.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{3}\). Here (OC-2=\(\sqrt{2}\)2+12=3), so \(OC=\sqrt{3}\). At each new step, (1) is added to the previous square under the root.
Step 3
Exam Tip
यहां (OC-2=\(\sqrt{2}\)2+12=3), इसलिए \(OC=\sqrt{3}\)। हर नए चरण में पिछले वर्गमूल में (1) जोड़ा जाता है।
On \(OC=\sqrt{3}\), drawing a perpendicular of (1) unit gives \(OD^2=3+1=4\). Remembering the order is very useful in such questions.
Step 2
Why this answer is correct
The correct answer is C. (OC). On \(OC=\sqrt{3}\), drawing a perpendicular of (1) unit gives \(OD^2=3+1=4\). Remembering the order is very useful in such questions.
Step 3
Exam Tip
\(OC=\sqrt{3}\) के ऊपर (1) इकाई का लंब बनाने पर \(OD^2=3+1=4\)। क्रम याद रखना ऐसे प्रश्नों में सबसे उपयोगी है।
The square of the new radius is (8+1=9), so the length is \(\sqrt{9}\). Remember that the length is written as \(\sqrt{9}\), not just (9).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{9}\). The square of the new radius is (8+1=9), so the length is \(\sqrt{9}\). Remember that the length is written as \(\sqrt{9}\), not just (9).
Step 3
Exam Tip
नई त्रिज्या का वर्ग (8+1=9) होगा, इसलिए लंबाई \(\sqrt{9}\) है। ध्यान रखें कि लंबाई (9) नहीं बल्कि \(\sqrt{9}\) लिखी जाती है।
B. \(\sqrt{6}\) पर (1) इकाई लंब बनाकर/By drawing (1) unit perpendicular on \(\sqrt{6}\)
Step 1
Concept
Because (\(\sqrt{6}\)2+12=7), the new segment becomes \(\sqrt{7}\). Identifying the correct previous segment is necessary.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{6}\) पर (1) इकाई लंब बनाकर / By drawing (1) unit perpendicular on \(\sqrt{6}\). Because (\(\sqrt{6}\)2+12=7), the new segment becomes \(\sqrt{7}\). Identifying the correct previous segment is necessary.
Step 3
Exam Tip
क्योंकि (\(\sqrt{6}\)2+12=7), इसलिए नया खंड \(\sqrt{7}\) होगा। सही पिछले खंड को पहचानना जरूरी है।
Putting (n=12) in the formula gives \(OP_{12}=\sqrt{13}\). Pay attention to the difference between the index and the number under the root.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{13}\). Putting (n=12) in the formula gives \(OP_{12}=\sqrt{13}\). Pay attention to the difference between the index and the number under the root.
Step 3
Exam Tip
सूत्र में (n=12) रखने पर \(OP_{12}=\sqrt{13}\) मिलता है। सूचकांक और वर्गमूल के अंदर की संख्या में अंतर पर ध्यान दें।
From \(\sqrt{2}\) to \(\sqrt{10}\), the number under the root increases by (8), so (8) new steps are needed. Each step increases the inner number by (1).
Step 2
Why this answer is correct
The correct answer is C. (8). From \(\sqrt{2}\) to \(\sqrt{10}\), the number under the root increases by (8), so (8) new steps are needed. Each step increases the inner number by (1).
Step 3
Exam Tip
\(\sqrt{2}\) से \(\sqrt{10}\) तक वर्गमूल के अंदर संख्या (8) बढ़ती है, इसलिए (8) नए चरण चाहिए। हर चरण अंदर की संख्या को (1) बढ़ाता है।
In every new right triangle, a perpendicular side of (1) unit is added. This is why the next hypotenuse represents the next square root.
Step 2
Why this answer is correct
The correct answer is B. (1). In every new right triangle, a perpendicular side of (1) unit is added. This is why the next hypotenuse represents the next square root.
Step 3
Exam Tip
हर नए समकोण त्रिभुज में (1) इकाई की लंब भुजा जोड़ी जाती है। इसी कारण अगला कर्ण अगले वर्गमूल को दर्शाता है।
B. हर नए कर्ण का वर्ग पिछले कर्ण के वर्ग से (1) अधिक होता है/The square of each new hypotenuse is (1) more than the square of the previous hypotenuse
Step 1
Concept
In the square root spiral, the length itself does not increase by (1); its square increases by (1). This difference helps solve many difficult questions.
Step 2
Why this answer is correct
The correct answer is B. हर नए कर्ण का वर्ग पिछले कर्ण के वर्ग से (1) अधिक होता है / The square of each new hypotenuse is (1) more than the square of the previous hypotenuse. In the square root spiral, the length itself does not increase by (1); its square increases by (1). This difference helps solve many difficult questions.
Step 3
Exam Tip
वर्गमूल सर्पिल में लंबाई नहीं बल्कि लंबाई का वर्ग (1) से बढ़ता है। इसी फर्क से कई कठिन प्रश्न हल होते हैं।
The order is \(\sqrt{2},\sqrt{3},\sqrt{4},\sqrt{5}\). Therefore, \(\sqrt{5}\) comes immediately after \(\sqrt{4}\).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{4}\). The order is \(\sqrt{2},\sqrt{3},\sqrt{4},\sqrt{5}\). Therefore, \(\sqrt{5}\) comes immediately after \(\sqrt{4}\).
Step 3
Exam Tip
क्रम \(\sqrt{2},\sqrt{3},\sqrt{4},\sqrt{5}\) चलता है। इसलिए \(\sqrt{5}\), \(\sqrt{4}\) के तुरंत बाद आता है।
B. समकोण त्रिभुज और पाइथागोरस प्रमेय का प्रयोग होता है/Right triangles and Pythagoras theorem are used
Step 1
Concept
The whole construction of the spiral is based on right triangles. Pythagoras theorem gives the length of the new hypotenuse.
Step 2
Why this answer is correct
The correct answer is B. समकोण त्रिभुज और पाइथागोरस प्रमेय का प्रयोग होता है / Right triangles and Pythagoras theorem are used. The whole construction of the spiral is based on right triangles. Pythagoras theorem gives the length of the new hypotenuse.
Step 3
Exam Tip
सर्पिल की पूरी रचना समकोण त्रिभुजों पर आधारित है। पाइथागोरस प्रमेय से नए कर्ण की लंबाई मिलती है।
The first triangle gives \(\sqrt{2}\), the second \(\sqrt{3}\), the third \(\sqrt{4}\), and the fourth \(\sqrt{5}\). Add (1) to the step number.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{5}\). The first triangle gives \(\sqrt{2}\), the second \(\sqrt{3}\), the third \(\sqrt{4}\), and the fourth \(\sqrt{5}\). Add (1) to the step number.
Step 3
Exam Tip
पहला त्रिभुज \(\sqrt{2}\), दूसरा \(\sqrt{3}\), तीसरा \(\sqrt{4}\), चौथा \(\sqrt{5}\) देता है। चरण संख्या में (1) जोड़ें।
(\(\sqrt{12}\)2+12=13), so the next perpendicular is drawn on the hypotenuse \(\sqrt{12}\). Identify the previous hypotenuse by reducing the inner number by (1).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{12}\). (\(\sqrt{12}\)2+12=13), so the next perpendicular is drawn on the hypotenuse \(\sqrt{12}\). Identify the previous hypotenuse by reducing the inner number by (1).
Step 3
Exam Tip
(\(\sqrt{12}\)2+12=13), इसलिए \(\sqrt{12}\) वाले कर्ण पर अगला लंब बनेगा। पिछले कर्ण को एक कम अंदर संख्या से पहचानें।
In the next step, a perpendicular of (1) unit is added, so the new distance becomes \(\sqrt{17+1}=\sqrt{18}\). The number under the root increases.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{18}\). In the next step, a perpendicular of (1) unit is added, so the new distance becomes \(\sqrt{17+1}=\sqrt{18}\). The number under the root increases.
Step 3
Exam Tip
अगले चरण में (1) इकाई लंब जुड़ता है, इसलिए नई दूरी \(\sqrt{17+1}=\sqrt{18}\) होगी। वर्गमूल के अंदर की संख्या बढ़ती है।
For consecutive hypotenuses, the numbers under the roots are consecutive. Therefore, \(\sqrt{9}\) and \(\sqrt{10}\) form the correct pair.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{9},\sqrt{10}\). For consecutive hypotenuses, the numbers under the roots are consecutive. Therefore, \(\sqrt{9}\) and \(\sqrt{10}\) form the correct pair.
Step 3
Exam Tip
लगातार कर्णों में वर्गमूल के अंदर की संख्याएं लगातार होती हैं। इसलिए \(\sqrt{9}\) और \(\sqrt{10}\) सही युग्म है।
In the immediately previous step, the number under the root is (1) less, so the length is \(\sqrt{19}\). Learn to read the sequence backward too.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{19}\). In the immediately previous step, the number under the root is (1) less, so the length is \(\sqrt{19}\). Learn to read the sequence backward too.
Step 3
Exam Tip
ठीक पहले वाले चरण में वर्गमूल के अंदर की संख्या (1) कम होगी, इसलिए लंबाई \(\sqrt{19}\) है। क्रम को पीछे से पढ़ना भी सीखें।
B. दोनों परिमेय संख्याएं हैं/Both are rational numbers
Step 1
Concept
\(\sqrt{4}=2\) and \(\sqrt{9}=3\), so both are rational. The spiral can show both rational and irrational square roots.
Step 2
Why this answer is correct
The correct answer is B. दोनों परिमेय संख्याएं हैं / Both are rational numbers. \(\sqrt{4}=2\) and \(\sqrt{9}=3\), so both are rational. The spiral can show both rational and irrational square roots.
Step 3
Exam Tip
\(\sqrt{4}=2\) और \(\sqrt{9}=3\), इसलिए दोनों परिमेय हैं। सर्पिल परिमेय और अपरिमेय दोनों प्रकार के वर्गमूल दिखा सकता है।
\(\sqrt{2}\) is irrational, so its decimal value does not terminate. The spiral represents it exactly geometrically.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}\). \(\sqrt{2}\) is irrational, so its decimal value does not terminate. The spiral represents it exactly geometrically.
Step 3
Exam Tip
\(\sqrt{2}\) अपरिमेय है, इसलिए इसका दशमलव मान समाप्त नहीं होता। सर्पिल इसे ज्यामितीय रूप से ठीक दर्शाता है।
Then the hypotenuse is \(\sqrt{2^2+1^2}=\sqrt{5}\), so the standard spiral is not formed. It is necessary to start with (OA=1).
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{5}\). Then the hypotenuse is \(\sqrt{2^2+1^2}=\sqrt{5}\), so the standard spiral is not formed. It is necessary to start with (OA=1).
Step 3
Exam Tip
तब कर्ण \(\sqrt{2^2+1^2}=\sqrt{5}\) होगा, इसलिए मानक सर्पिल नहीं बनेगा। शुरुआत में (OA=1) लेना जरूरी है।
A. वह पिछले कर्ण पर लंब होना चाहिए/It must be perpendicular to the previous hypotenuse
Step 1
Concept
Every new (1) unit segment is drawn perpendicular to the previous hypotenuse. This is the correct geometric construction of the spiral.
Step 2
Why this answer is correct
The correct answer is A. वह पिछले कर्ण पर लंब होना चाहिए / It must be perpendicular to the previous hypotenuse. Every new (1) unit segment is drawn perpendicular to the previous hypotenuse. This is the correct geometric construction of the spiral.
Step 3
Exam Tip
हर नया (1) इकाई खंड पिछले कर्ण पर लंब बनाया जाता है। यही सर्पिल की सही ज्यामितीय रचना है।
B. \(\sqrt{7}\) पर (1) इकाई लंब बनाना/Draw a perpendicular of (1) unit on \(\sqrt{7}\)
Step 1
Concept
(\(\sqrt{7}\)2+12=8), so the new hypotenuse becomes \(\sqrt{8}\). The process always starts from the previous square root.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{7}\) पर (1) इकाई लंब बनाना / Draw a perpendicular of (1) unit on \(\sqrt{7}\). (\(\sqrt{7}\)2+12=8), so the new hypotenuse becomes \(\sqrt{8}\). The process always starts from the previous square root.
Step 3
Exam Tip
(\(\sqrt{7}\)2+12=8), इसलिए नया कर्ण \(\sqrt{8}\) बनेगा। प्रक्रिया हमेशा पिछले वर्गमूल से शुरू होती है।
Because the new (1) unit segment is perpendicular to the previous hypotenuse, \(OQ^2=OP^2+1\). Keep squares and lengths separate while writing the formula.
Step 2
Why this answer is correct
The correct answer is A. \(OQ^2=OP^2+1\). Because the new (1) unit segment is perpendicular to the previous hypotenuse, \(OQ^2=OP^2+1\). Keep squares and lengths separate while writing the formula.
Step 3
Exam Tip
क्योंकि नया (1) इकाई खंड पिछले कर्ण पर लंब होता है, इसलिए \(OQ^2=OP^2+1\)। सूत्र लिखते समय वर्ग और लंबाई अलग रखें।
The first triangle gives \(\sqrt{2}\), so for \(\sqrt{12}\), the order is (12-1=11). Subtract (1) from the inner number.
Step 2
Why this answer is correct
The correct answer is C. (11)वां / (11)th. The first triangle gives \(\sqrt{2}\), so for \(\sqrt{12}\), the order is (12-1=11). Subtract (1) from the inner number.
Step 3
Exam Tip
पहला त्रिभुज \(\sqrt{2}\) देता है, इसलिए \(\sqrt{12}\) के लिए क्रम (12-1=11) होगा। अंदर की संख्या से (1) घटाएं।
The hypotenuse of the (r)th triangle is \(\sqrt{r+1}\). Therefore, the (10)th triangle has hypotenuse \(\sqrt{11}\).
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{11}\). The hypotenuse of the (r)th triangle is \(\sqrt{r+1}\). Therefore, the (10)th triangle has hypotenuse \(\sqrt{11}\).
Step 3
Exam Tip
(r)वें त्रिभुज का कर्ण \(\sqrt{r+1}\) होता है। इसलिए (10)वें त्रिभुज का कर्ण \(\sqrt{11}\) है।
The start is (OA=1), and then the hypotenuses \(\sqrt{2},\sqrt{3},\sqrt{4}\) are formed. The number under the root increases continuously.
Step 2
Why this answer is correct
The correct answer is A. \(1,\sqrt{2},\sqrt{3},\sqrt{4}\). The start is (OA=1), and then the hypotenuses \(\sqrt{2},\sqrt{3},\sqrt{4}\) are formed. The number under the root increases continuously.
Step 3
Exam Tip
आरंभ (OA=1) से होता है और फिर कर्ण \(\sqrt{2},\sqrt{3},\sqrt{4}\) बनते हैं। क्रम में अंदर की संख्या लगातार बढ़ती है।
\(\sqrt{27}\) is formed by the (26)th triangle, so (26) new perpendicular segments of (1) unit are used. The first (1) unit perpendicular forms \(\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is B. (26). \(\sqrt{27}\) is formed by the (26)th triangle, so (26) new perpendicular segments of (1) unit are used. The first (1) unit perpendicular forms \(\sqrt{2}\).
Step 3
Exam Tip
\(\sqrt{27}\) (26)वें त्रिभुज से बनता है, इसलिए (26) नए (1) इकाई लंब खंड लगते हैं। पहला (1) लंब \(\sqrt{2}\) बनाता है।
For length (3), the hypotenuse must be \(\sqrt{9}\). Therefore, it is obtained at the \(\sqrt{9}\) step.
Step 2
Why this answer is correct
The correct answer is D. \(\sqrt{9}\) वाला चरण / Step of \(\sqrt{9}\). For length (3), the hypotenuse must be \(\sqrt{9}\). Therefore, it is obtained at the \(\sqrt{9}\) step.
Step 3
Exam Tip
लंबाई (3) के लिए कर्ण \(\sqrt{9}\) होना चाहिए। इसलिए यह \(\sqrt{9}\) वाले चरण पर मिलता है।
C. \(\sqrt{14}\) के बाद \(\sqrt{16}\) बनाना/Constructing \(\sqrt{16}\) after \(\sqrt{14}\)
Step 1
Concept
In one step, the number under the root increases only by (1), so \(\sqrt{15}\) should come after \(\sqrt{14}\). Skipping a step is incorrect.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{14}\) के बाद \(\sqrt{16}\) बनाना / Constructing \(\sqrt{16}\) after \(\sqrt{14}\). In one step, the number under the root increases only by (1), so \(\sqrt{15}\) should come after \(\sqrt{14}\). Skipping a step is incorrect.
Step 3
Exam Tip
एक चरण में अंदर की संख्या केवल (1) बढ़ती है, इसलिए \(\sqrt{14}\) के बाद \(\sqrt{15}\) आना चाहिए। बीच का चरण छोड़ना गलत है।
A. खंड समानांतर नहीं, लंब होना चाहिए/The segment should be perpendicular, not parallel
Step 1
Concept
In a square root spiral, the new (1) unit segment is drawn perpendicular to the previous hypotenuse. A parallel segment will not allow Pythagoras theorem to apply.
Step 2
Why this answer is correct
The correct answer is A. खंड समानांतर नहीं, लंब होना चाहिए / The segment should be perpendicular, not parallel. In a square root spiral, the new (1) unit segment is drawn perpendicular to the previous hypotenuse. A parallel segment will not allow Pythagoras theorem to apply.
Step 3
Exam Tip
वर्गमूल सर्पिल में नया (1) इकाई खंड पिछले कर्ण पर लंब बनाया जाता है। समानांतर खंड से पाइथागोरस प्रमेय लागू नहीं होगा।
This relation matches Pythagoras theorem, so (OP) and (PQ) are perpendicular. This right-angle construction is essential in the square root spiral.
Step 2
Why this answer is correct
The correct answer is B. समकोण / Right angle. This relation matches Pythagoras theorem, so (OP) and (PQ) are perpendicular. This right-angle construction is essential in the square root spiral.
Step 3
Exam Tip
यह संबंध पाइथागोरस प्रमेय जैसा है, इसलिए (OP) और (PQ) लंब होंगे। वर्गमूल सर्पिल में यही समकोण रचना जरूरी है।
A. क्रम \(\sqrt{1},\sqrt{2},\sqrt{3}\) स्पष्ट हो जाता है/The order \(\sqrt{1},\sqrt{2},\sqrt{3}\) becomes clear
Step 1
Concept
Taking \(1=\sqrt{1}\) makes the whole sequence clear. Then the inner number increases by (1) at each step.
Step 2
Why this answer is correct
The correct answer is A. क्रम \(\sqrt{1},\sqrt{2},\sqrt{3}\) स्पष्ट हो जाता है / The order \(\sqrt{1},\sqrt{2},\sqrt{3}\) becomes clear. Taking \(1=\sqrt{1}\) makes the whole sequence clear. Then the inner number increases by (1) at each step.
Step 3
Exam Tip
\(1=\sqrt{1}\) मानने से पूरा क्रम स्पष्ट दिखता है। फिर हर नए चरण में अंदर की संख्या (1) बढ़ती है।
B. वर्गों में बराबर (1) की वृद्धि/Equal increase of (1) in squares
Step 1
Concept
The squares of the hypotenuses are (2,3,4), which increase by (1). The actual lengths do not have equal differences.
Step 2
Why this answer is correct
The correct answer is B. वर्गों में बराबर (1) की वृद्धि / Equal increase of (1) in squares. The squares of the hypotenuses are (2,3,4), which increase by (1). The actual lengths do not have equal differences.
Step 3
Exam Tip
कर्णों के वर्ग (2,3,4) हैं, जिनमें (1) की वृद्धि है। वास्तविक लंबाइयों में समान अंतर नहीं होता।
The exact value is \(\sqrt{5}\), while (2.236) is only an approximation. In exams, prefer the exact square root form.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{5}\). The exact value is \(\sqrt{5}\), while (2.236) is only an approximation. In exams, prefer the exact square root form.
Step 3
Exam Tip
सटीक मान \(\sqrt{5}\) ही है, जबकि (2.236) केवल अनुमान है। परीक्षा में सटीक वर्गमूल रूप को प्राथमिकता दें।
\(\sqrt{6}\) is formed from the previous hypotenuse \(\sqrt{5}\) and the new perpendicular side (1). Therefore, (\(\sqrt{5}\)2+12=6) is correct.
Step 2
Why this answer is correct
The correct answer is A. (\(\sqrt{5}\)2+12=6). \(\sqrt{6}\) is formed from the previous hypotenuse \(\sqrt{5}\) and the new perpendicular side (1). Therefore, (\(\sqrt{5}\)2+12=6) is correct.
Step 3
Exam Tip
\(\sqrt{6}\) पिछले कर्ण \(\sqrt{5}\) और नई लंब भुजा (1) से बनता है। इसलिए संबंध (\(\sqrt{5}\)2+12=6) सही है।
\(\sqrt{2}\) is given by the first triangle, so up to \(\sqrt{n}\), the number of triangles is (n-1). Remember the starting term while writing a general formula.
Step 2
Why this answer is correct
The correct answer is B. (n-1). \(\sqrt{2}\) is given by the first triangle, so up to \(\sqrt{n}\), the number of triangles is (n-1). Remember the starting term while writing a general formula.
Step 3
Exam Tip
\(\sqrt{2}\) पहला त्रिभुज देता है, इसलिए \(\sqrt{n}\) तक त्रिभुजों की संख्या (n-1) होगी। सामान्य सूत्र लिखते समय आरंभिक पद याद रखें।
The first hypotenuse is formed from (1) and (1), giving \(\sqrt{2}\). \(\sqrt{4}\) appears in a later step, not as the first hypotenuse.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{4}\). The first hypotenuse is formed from (1) and (1), giving \(\sqrt{2}\). \(\sqrt{4}\) appears in a later step, not as the first hypotenuse.
Step 3
Exam Tip
पहला कर्ण (1) और (1) से बनकर \(\sqrt{2}\) होता है। \(\sqrt{4}\) बाद के चरण में मिलता है, पहले कर्ण के रूप में नहीं।
A. वर्गमूलों को ज्यामितीय रूप से निरूपित करना/To represent square roots geometrically
Step 1
Concept
The square root spiral represents square roots like \(\sqrt{2},\sqrt{3},\sqrt{5}\) geometrically. It helps in understanding the number line and irrational numbers.
Step 2
Why this answer is correct
The correct answer is A. वर्गमूलों को ज्यामितीय रूप से निरूपित करना / To represent square roots geometrically. The square root spiral represents square roots like \(\sqrt{2},\sqrt{3},\sqrt{5}\) geometrically. It helps in understanding the number line and irrational numbers.
Step 3
Exam Tip
वर्गमूल सर्पिल \(\sqrt{2},\sqrt{3},\sqrt{5}\) जैसे वर्गमूलों को ज्यामितीय रूप से दिखाता है। यह संख्या रेखा और अपरिमेय संख्याओं को समझने में मदद करता है।
