The square of the new hypotenuse is (35+1=36), so \(OQ=\sqrt{36}\). When the hypotenuse is asked, write the square root form.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{36}\). The square of the new hypotenuse is (35+1=36), so \(OQ=\sqrt{36}\). When the hypotenuse is asked, write the square root form.
Step 3
Exam Tip
नए कर्ण का वर्ग (35+1=36) होगा, इसलिए \(OQ=\sqrt{36}\)। कर्ण पूछने पर वर्गमूल रूप लिखें।
In the previous step, the number under the root is (1) less. Therefore, \(\sqrt{41}\) comes before \(\sqrt{42}\).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{41}\). In the previous step, the number under the root is (1) less. Therefore, \(\sqrt{41}\) comes before \(\sqrt{42}\).
Step 3
Exam Tip
पिछले चरण में वर्गमूल के अंदर की संख्या (1) कम होती है। इसलिए \(\sqrt{42}\) से पहले \(\sqrt{41}\) होगा।
The hypotenuse of the (r)th triangle is \(\sqrt{r+1}\). Therefore, the (18)th triangle has hypotenuse \(\sqrt{19}\).
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{19}\). The hypotenuse of the (r)th triangle is \(\sqrt{r+1}\). Therefore, the (18)th triangle has hypotenuse \(\sqrt{19}\).
Step 3
Exam Tip
(r)वें त्रिभुज का कर्ण \(\sqrt{r+1}\) होता है। इसलिए (18)वें त्रिभुज का कर्ण \(\sqrt{19}\) है।
\(\sqrt{2}\) is given by the first triangle, so up to \(\sqrt{48}\), the number of triangles is (48-1=47). Keep the initial \(\sqrt{1}\) separate in counting.
Step 2
Why this answer is correct
The correct answer is B. (47). \(\sqrt{2}\) is given by the first triangle, so up to \(\sqrt{48}\), the number of triangles is (48-1=47). Keep the initial \(\sqrt{1}\) separate in counting.
Step 3
Exam Tip
\(\sqrt{2}\) पहला त्रिभुज देता है, इसलिए \(\sqrt{48}\) तक त्रिभुजों की संख्या (48-1=47) होगी। गणना में आरंभिक \(\sqrt{1}\) को अलग रखें।
In consecutive hypotenuses, the numbers under the roots increase by (1). Therefore, \(\sqrt{22},\sqrt{23},\sqrt{24}\) is the correct order.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{22},\sqrt{23},\sqrt{24}\). In consecutive hypotenuses, the numbers under the roots increase by (1). Therefore, \(\sqrt{22},\sqrt{23},\sqrt{24}\) is the correct order.
Step 3
Exam Tip
लगातार कर्णों में वर्गमूल के अंदर की संख्याएं (1) के अंतर से बढ़ती हैं। इसलिए \(\sqrt{22},\sqrt{23},\sqrt{24}\) सही क्रम है।
(\(\sqrt{30}\)2+12=31), so the perpendicular is drawn on \(\sqrt{30}\). For the next square root, identify the previous inner value.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{30}\). (\(\sqrt{30}\)2+12=31), so the perpendicular is drawn on \(\sqrt{30}\). For the next square root, identify the previous inner value.
Step 3
Exam Tip
(\(\sqrt{30}\)2+12=31), इसलिए \(\sqrt{30}\) पर लंब बनता है। अगले वर्गमूल के लिए पिछले अंदर मान को पहचानें।
A. लंबाइयों का अंतर (1) नहीं, उनके वर्गों का अंतर (1) है/The difference of lengths is not (1), the difference of their squares is (1)
Step 1
Concept
(\(\sqrt{11}\)2-\(\sqrt{10}\)2=1), but \(\sqrt{11}-\sqrt{10}\neq1\). In exams, understand the difference between length and square.
Step 2
Why this answer is correct
The correct answer is A. लंबाइयों का अंतर (1) नहीं, उनके वर्गों का अंतर (1) है / The difference of lengths is not (1), the difference of their squares is (1). (\(\sqrt{11}\)2-\(\sqrt{10}\)2=1), but \(\sqrt{11}-\sqrt{10}\neq1\). In exams, understand the difference between length and square.
Step 3
Exam Tip
(\(\sqrt{11}\)2-\(\sqrt{10}\)2=1), लेकिन \(\sqrt{11}-\sqrt{10}\neq1\)। परीक्षा में लंबाई और वर्ग में अंतर समझें।
\(\sqrt{50}=5\sqrt{2}\), and \(\sqrt{2}\) is irrational. Therefore, \(\sqrt{50}\) is an irrational number.
Step 2
Why this answer is correct
The correct answer is C. अपरिमेय संख्या / Irrational number. \(\sqrt{50}=5\sqrt{2}\), and \(\sqrt{2}\) is irrational. Therefore, \(\sqrt{50}\) is an irrational number.
Step 3
Exam Tip
\(\sqrt{50}=5\sqrt{2}\) है और \(\sqrt{2}\) अपरिमेय है। इसलिए \(\sqrt{50}\) अपरिमेय संख्या है।
In the next step, the inner number increases by (1), so \(\sqrt{81}\) is formed. The spiral does not skip numbers in order.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{81}\). In the next step, the inner number increases by (1), so \(\sqrt{81}\) is formed. The spiral does not skip numbers in order.
Step 3
Exam Tip
अगले चरण में अंदर की संख्या (1) बढ़ती है, इसलिए \(\sqrt{81}\) बनेगा। सर्पिल क्रम में कोई संख्या नहीं छोड़ता।
B. \(\sqrt{98}\) और \(\sqrt{100}\)/\(\sqrt{98}\) and \(\sqrt{100}\)
Step 1
Concept
In order, the inner number decreases by (1) before and increases by (1) after. So \(\sqrt{98}\) comes before \(\sqrt{99}\), and \(\sqrt{100}\) comes after it.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{98}\) और \(\sqrt{100}\) / \(\sqrt{98}\) and \(\sqrt{100}\). In order, the inner number decreases by (1) before and increases by (1) after. So \(\sqrt{98}\) comes before \(\sqrt{99}\), and \(\sqrt{100}\) comes after it.
Step 3
Exam Tip
क्रम में अंदर की संख्या (1) घटती और (1) बढ़ती है। इसलिए \(\sqrt{99}\) के पहले \(\sqrt{98}\) और बाद में \(\sqrt{100}\) होगा।
B. पिछले कर्ण पर (1) इकाई का लंब बनाना/Drawing a (1) unit perpendicular to the previous hypotenuse
Step 1
Concept
The main construction depends on drawing a (1) unit perpendicular to the previous hypotenuse. This creates the next right triangle.
Step 2
Why this answer is correct
The correct answer is B. पिछले कर्ण पर (1) इकाई का लंब बनाना / Drawing a (1) unit perpendicular to the previous hypotenuse. The main construction depends on drawing a (1) unit perpendicular to the previous hypotenuse. This creates the next right triangle.
Step 3
Exam Tip
मुख्य रचना पिछले कर्ण पर (1) इकाई का लंब बनाने पर निर्भर करती है। इसी से अगला समकोण त्रिभुज बनता है।
B. नहीं, दिशा बदलती है क्योंकि प्रत्येक नया लंब पिछले कर्ण पर बनता है/No, the direction changes because each new perpendicular is drawn on the previous hypotenuse
Step 1
Concept
At each new step, the perpendicular is drawn on the previous hypotenuse, so the direction gradually changes. The spiral is formed by this turning.
Step 2
Why this answer is correct
The correct answer is B. नहीं, दिशा बदलती है क्योंकि प्रत्येक नया लंब पिछले कर्ण पर बनता है / No, the direction changes because each new perpendicular is drawn on the previous hypotenuse. At each new step, the perpendicular is drawn on the previous hypotenuse, so the direction gradually changes. The spiral is formed by this turning.
Step 3
Exam Tip
हर नए चरण में लंब पिछले कर्ण पर बनता है, इसलिए दिशा धीरे-धीरे बदलती है। सर्पिल इसी घुमाव से बनता है।
B. नहीं, क्योंकि पाइथागोरस प्रमेय के लिए समकोण जरूरी है/No, because a right angle is necessary for Pythagoras theorem
Step 1
Concept
\(\sqrt{16}\) is certain only when the new segment is perpendicular to the previous hypotenuse. Without a right angle, Pythagoras theorem cannot be applied.
Step 2
Why this answer is correct
The correct answer is B. नहीं, क्योंकि पाइथागोरस प्रमेय के लिए समकोण जरूरी है / No, because a right angle is necessary for Pythagoras theorem. \(\sqrt{16}\) is certain only when the new segment is perpendicular to the previous hypotenuse. Without a right angle, Pythagoras theorem cannot be applied.
Step 3
Exam Tip
\(\sqrt{16}\) तभी निश्चित होगा जब नया खंड पिछले कर्ण पर लंब हो। समकोण के बिना पाइथागोरस प्रमेय नहीं लगेगा।
The next hypotenuse is \(\sqrt{101}\), which is not a simple integer. Even after \(\sqrt{100}=10\), the sequence continues with \(\sqrt{101}\).
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{101}\). The next hypotenuse is \(\sqrt{101}\), which is not a simple integer. Even after \(\sqrt{100}=10\), the sequence continues with \(\sqrt{101}\).
Step 3
Exam Tip
अगला कर्ण \(\sqrt{101}\) होगा, जो सरल पूर्णांक नहीं है। \(\sqrt{100}=10\) के बाद भी क्रम \(\sqrt{101}\) से चलता है।
A. सर्पिल से मिली दूरी को कंपास से संख्या रेखा पर चिह्नित किया जा सकता है/The distance obtained from the spiral can be marked on the number line using a compass
Step 1
Concept
The spiral gives an exact distance that can be transferred to the number line with a compass. It is useful for showing irrational numbers.
Step 2
Why this answer is correct
The correct answer is A. सर्पिल से मिली दूरी को कंपास से संख्या रेखा पर चिह्नित किया जा सकता है / The distance obtained from the spiral can be marked on the number line using a compass. The spiral gives an exact distance that can be transferred to the number line with a compass. It is useful for showing irrational numbers.
Step 3
Exam Tip
सर्पिल सटीक दूरी देता है, जिसे कंपास से संख्या रेखा पर स्थानांतरित किया जा सकता है। यह अपरिमेय संख्याओं को दिखाने में उपयोगी है।
C. हर नए लंब खंड की लंबाई (2)/Length of every new perpendicular segment is (2)
Step 1
Concept
In the standard spiral, every new perpendicular segment is (1) unit long. Taking (2) units would create a different sequence.
Step 2
Why this answer is correct
The correct answer is C. हर नए लंब खंड की लंबाई (2) / Length of every new perpendicular segment is (2). In the standard spiral, every new perpendicular segment is (1) unit long. Taking (2) units would create a different sequence.
Step 3
Exam Tip
मानक सर्पिल में हर नया लंब खंड (1) इकाई का होता है। (2) इकाई लेने से अलग क्रम बनेगा।
\(\sqrt{120}=\sqrt{4\times30}=2\sqrt{30}\) is simplest, while \(4\sqrt{15}\) squares to (240). Check each option carefully.
Step 2
Why this answer is correct
The correct answer is B. \(4\sqrt{15}\). \(\sqrt{120}=\sqrt{4\times30}=2\sqrt{30}\) is simplest, while \(4\sqrt{15}\) squares to (240). Check each option carefully.
Step 3
Exam Tip
\(\sqrt{120}=\sqrt{4\times30}=2\sqrt{30}\) भी सही है, लेकिन \(\sqrt{120}=\sqrt{4\times30}\) से और बड़ा पूर्ण वर्ग नहीं निकलता। विकल्पों में \(2\sqrt{30}\) नहीं सही दिया गया है, इसलिए ध्यान से जांचें।
\(\sqrt{2}\) is the first hypotenuse, so the (25)th hypotenuse is \(\sqrt{25+1}=\sqrt{26}\). Keep the starting point of counting clear.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{26}\). \(\sqrt{2}\) is the first hypotenuse, so the (25)th hypotenuse is \(\sqrt{25+1}=\sqrt{26}\). Keep the starting point of counting clear.
Step 3
Exam Tip
\(\sqrt{2}\) पहला कर्ण है, इसलिए (25)वां कर्ण \(\sqrt{25+1}=\sqrt{26}\) होगा। गिनती का आरंभ स्पष्ट रखें।
(\(\sqrt{12}\)2+12=13), so the result will be \(\sqrt{13}\). To get \(\sqrt{14}\), \(\sqrt{13}\) is needed first.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{13}\). (\(\sqrt{12}\)2+12=13), so the result will be \(\sqrt{13}\). To get \(\sqrt{14}\), \(\sqrt{13}\) is needed first.
Step 3
Exam Tip
(\(\sqrt{12}\)2+12=13), इसलिए परिणाम \(\sqrt{13}\) होगा। \(\sqrt{14}\) के लिए पहले \(\sqrt{13}\) चाहिए।
B. \(\sqrt{144}\) परिमेय है और \(\sqrt{145}\) अपरिमेय है/\(\sqrt{144}\) is rational and \(\sqrt{145}\) is irrational
Step 1
Concept
\(\sqrt{144}=12\) is rational, while (145) is not a perfect square. Therefore, \(\sqrt{145}\) is irrational.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{144}\) परिमेय है और \(\sqrt{145}\) अपरिमेय है / \(\sqrt{144}\) is rational and \(\sqrt{145}\) is irrational. \(\sqrt{144}=12\) is rational, while (145) is not a perfect square. Therefore, \(\sqrt{145}\) is irrational.
Step 3
Exam Tip
\(\sqrt{144}=12\) परिमेय है, जबकि (145) पूर्ण वर्ग नहीं है। इसलिए \(\sqrt{145}\) अपरिमेय है।
The first right triangle has perpendicular sides (1) and (1). 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 \(1^2+1^2=2\). The first right triangle has perpendicular sides (1) and (1). Therefore, the hypotenuse is \(\sqrt{2}\).
Step 3
Exam Tip
पहले समकोण त्रिभुज की दोनों लंब भुजाएं (1) और (1) होती हैं। इसलिए कर्ण \(\sqrt{2}\) होता है।
A. सर्पिल में मिली दूरी को कंपास से संख्या रेखा पर स्थानांतरित करें/Transfer the distance obtained in the spiral to the number line using a compass
Step 1
Concept
The spiral gives the exact geometric distance of \(\sqrt{6}\). The same distance can be placed on the number line using a compass.
Step 2
Why this answer is correct
The correct answer is A. सर्पिल में मिली दूरी को कंपास से संख्या रेखा पर स्थानांतरित करें / Transfer the distance obtained in the spiral to the number line using a compass. The spiral gives the exact geometric distance of \(\sqrt{6}\). The same distance can be placed on the number line using a compass.
Step 3
Exam Tip
सर्पिल \(\sqrt{6}\) की सटीक ज्यामितीय दूरी देता है। कंपास से वही दूरी संख्या रेखा पर रखी जा सकती है।
From \(\sqrt{2}\) to \(\sqrt{9}\), the inner numbers (2,3,4,5,6,7,8,9) give (8) hypotenuses. \(\sqrt{4}\) and \(\sqrt{9}\) also give integer lengths.
Step 2
Why this answer is correct
The correct answer is A. इसमें (8) कर्ण बनते हैं / It has (8) hypotenuses. From \(\sqrt{2}\) to \(\sqrt{9}\), the inner numbers (2,3,4,5,6,7,8,9) give (8) hypotenuses. \(\sqrt{4}\) and \(\sqrt{9}\) also give integer lengths.
Step 3
Exam Tip
\(\sqrt{2}\) से \(\sqrt{9}\) तक (2,3,4,5,6,7,8,9) कुल (8) कर्ण हैं। \(\sqrt{4}\) और \(\sqrt{9}\) पूर्णांक लंबाइयां भी देते हैं।
The new hypotenuse is \(\sqrt{18+1}=\sqrt{19}\). Since (19) is not a perfect square, this is already the simplest form.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{19}\). The new hypotenuse is \(\sqrt{18+1}=\sqrt{19}\). Since (19) is not a perfect square, this is already the simplest form.
Step 3
Exam Tip
नया कर्ण \(\sqrt{18+1}=\sqrt{19}\) होगा। (19) पूर्ण वर्ग नहीं है, इसलिए यही सरल रूप है।
\(\sqrt{4}=2\), which is a rational number. Not all square roots formed in the spiral are irrational.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि \(\sqrt{4}=2\) / Because \(\sqrt{4}=2\). \(\sqrt{4}=2\), which is a rational number. Not all square roots formed in the spiral are irrational.
Step 3
Exam Tip
\(\sqrt{4}=2\) है, जो परिमेय संख्या है। सर्पिल में बने सभी वर्गमूल अपरिमेय नहीं होते।
A. \(अगला कर्ण (=\sqrt{\)पिछले कर्ण का वर्ग+1})/\(Next hypotenuse (=\sqrt{\)square of previous hypotenuse+1})
Step 1
Concept
\(By Pythagoras theorem, the next hypotenuse is (\sqrt{(\)previous hypotenuse\()^2+1}). So both square and square root must be in the correct places.\)
Step 2
Why this answer is correct
\(The correct answer is A. अगला कर्ण (=\sqrt{\)पिछले कर्ण का वर्ग\(+1}) / Next hypotenuse (=\sqrt{\)square of previous hypotenuse\(+1}). By Pythagoras theorem, the next hypotenuse is (\sqrt{(\)previous hypotenuse\()^2+1}). So both square and square root must be in the correct places.\)
Step 3
Exam Tip
\(पाइथागोरस प्रमेय से अगला कर्ण (\sqrt{(\)पिछला कर्ण)2+1}) होता है। इसलिए वर्ग और वर्गमूल दोनों सही जगह होने चाहिए।
If the length is (5), its square is (25), so the hypotenuse appears as \(\sqrt{25}\). Connect integer lengths with their square root forms.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{25}\). If the length is (5), its square is (25), so the hypotenuse appears as \(\sqrt{25}\). Connect integer lengths with their square root forms.
Step 3
Exam Tip
लंबाई (5) होने पर वर्ग (25) होगा, इसलिए कर्ण \(\sqrt{25}\) के रूप में दिखेगा। पूर्णांक लंबाइयों को उनके वर्गमूल रूप से जोड़ें।
In a square root spiral, every new hypotenuse is formed using a right triangle. Therefore, its main basis is Pythagoras theorem.
Step 2
Why this answer is correct
The correct answer is B. पाइथागोरस प्रमेय / Pythagoras theorem. In a square root spiral, every new hypotenuse is formed using a right triangle. Therefore, its main basis is Pythagoras theorem.
Step 3
Exam Tip
वर्गमूल सर्पिल में हर नया कर्ण समकोण त्रिभुज से बनता है। इसलिए इसका मुख्य आधार पाइथागोरस प्रमेय है।
