वर्गमूल सर्पिल में सबसे पहले कौन-सा रेखाखंड बनाया जाता है?
Which line segment is drawn first in a square root spiral?
#square-root-spiral
#construction
#start
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A (1) इकाई का रेखाखंड / A line segment of (1) unit
B (2) इकाई का रेखाखंड / A line segment of (2) units
C \(\sqrt{2}\) इकाई का रेखाखंड / A line segment of \(\sqrt{2}\) units
D \(\sqrt{3}\) इकाई का रेखाखंड / A line segment of \(\sqrt{3}\) units
Explanation opens after your attempt
Correct Answer
A. (1) इकाई का रेखाखंड / A line segment of (1) unit
Step 1
Concept
The construction starts with a (1) unit line segment. The first right triangle is made from it.
Step 2
Why this answer is correct
The correct answer is A. (1) इकाई का रेखाखंड / A line segment of (1) unit. The construction starts with a (1) unit line segment. The first right triangle is made from it.
Step 3
Exam Tip
शुरुआत (1) इकाई के रेखाखंड से होती है। इसी से पहला समकोण त्रिभुज बनाया जाता है।
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वर्गमूल सर्पिल में (1) इकाई और (1) इकाई भुजाओं से बने समकोण त्रिभुज का कर्ण क्या होगा?
In a square root spiral, what is the hypotenuse of a right triangle with sides (1) unit and (1) unit?
#square-root-spiral
#pythagoras
#sqrt2
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A \(\sqrt{1}\)
B \(\sqrt{2}\)
C \(\sqrt{3}\)
D \(\sqrt{4}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{2}\)
Step 1
Concept
By Pythagoras theorem the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). The first main hypotenuse is \(\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{2}\). By Pythagoras theorem the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). The first main hypotenuse is \(\sqrt{2}\).
Step 3
Exam Tip
पाइथागोरस प्रमेय से कर्ण \(\sqrt{1^2+1^2}=\sqrt{2}\) होता है। पहला मुख्य कर्ण \(\sqrt{2}\) है।
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वर्गमूल सर्पिल में प्रत्येक नए त्रिभुज की नई लंब भुजा कितनी रखी जाती है?
In a square root spiral, what is the length of the new perpendicular side in each new triangle?
#square-root-spiral
#unit-perpendicular
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A \(\sqrt{2}\) इकाई / \(\sqrt{2}\) units
B (1) इकाई / (1) unit
C (2) इकाई / (2) units
D \(\sqrt{3}\) इकाई / \(\sqrt{3}\) units
Explanation opens after your attempt
Correct Answer
B. (1) इकाई / (1) unit
Step 1
Concept
At every step the new perpendicular side is kept (1) unit. This forms the next square root.
Step 2
Why this answer is correct
The correct answer is B. (1) इकाई / (1) unit. At every step the new perpendicular side is kept (1) unit. This forms the next square root.
Step 3
Exam Tip
हर चरण में नई लंब भुजा (1) इकाई रखी जाती है। इसी से अगला वर्गमूल बनता है।
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वर्गमूल सर्पिल में समकोण त्रिभुज का कर्ण निकालने के लिए कौन-सा प्रमेय उपयोग होता है?
Which theorem is used to find the hypotenuse of a right triangle in a square root spiral?
#square-root-spiral
#pythagoras
#theorem
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A थेल्स प्रमेय / Thales theorem
B मध्यबिंदु प्रमेय / Midpoint theorem
C पाइथागोरस प्रमेय / Pythagoras theorem
D शेषफल प्रमेय / Remainder theorem
Explanation opens after your attempt
Correct Answer
C. पाइथागोरस प्रमेय / Pythagoras theorem
Step 1
Concept
Pythagoras theorem is used to find the hypotenuse. So recognizing the right triangle is important.
Step 2
Why this answer is correct
The correct answer is C. पाइथागोरस प्रमेय / Pythagoras theorem. Pythagoras theorem is used to find the hypotenuse. So recognizing the right triangle is important.
Step 3
Exam Tip
कर्ण ज्ञात करने के लिए पाइथागोरस प्रमेय लगाया जाता है। इसलिए समकोण त्रिभुज पहचानना जरूरी है।
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\(\sqrt{2}\) कर्ण पर (1) इकाई लंब जोड़ने पर अगला कर्ण क्या बनेगा?
What will be the next hypotenuse after adding a (1) unit perpendicular to the hypotenuse \(\sqrt{2}\)?
#square-root-spiral
#next-root
#sqrt3
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A \(\sqrt{2}\)
B \(\sqrt{4}\)
C \(\sqrt{5}\)
D \(\sqrt{3}\)
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Correct Answer
D. \(\sqrt{3}\)
Step 1
Concept
The new hypotenuse is (\sqrt{\(\sqrt{2}\)2 +12 }=\sqrt{3}). Each time (1) is added to the number.
Step 2
Why this answer is correct
The correct answer is D. \(\sqrt{3}\). The new hypotenuse is (\sqrt{\(\sqrt{2}\)2 +12 }=\sqrt{3}). Each time (1) is added to the number.
Step 3
Exam Tip
नया कर्ण (\sqrt{\(\sqrt{2}\)2 +12 }=\sqrt{3}) होगा। हर बार संख्या में (1) जुड़ता है।
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\(\sqrt{3}\) के बाद वर्गमूल सर्पिल में अगला कर्ण कौन-सा होगा?
After \(\sqrt{3}\), which hypotenuse comes next in a square root spiral?
#square-root-spiral
#sequence
#next-root
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A \(\sqrt{2}\)
B \(\sqrt{4}\)
C \(\sqrt{5}\)
D \(\sqrt{6}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{4}\)
Step 1
Concept
With \(\sqrt{3}\) and a (1) unit perpendicular the new hypotenuse is \(\sqrt{4}\). Take the next square root in sequence.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{4}\). With \(\sqrt{3}\) and a (1) unit perpendicular the new hypotenuse is \(\sqrt{4}\). Take the next square root in sequence.
Step 3
Exam Tip
\(\sqrt{3}\) के साथ (1) इकाई लंब से नया कर्ण \(\sqrt{4}\) बनता है। क्रम में अगला वर्गमूल लें।
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वर्गमूल सर्पिल में कर्णों का सही प्रारंभिक क्रम कौन-सा है?
What is the correct initial sequence of hypotenuses in a square root spiral?
#square-root-spiral
#sequence
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A \(\sqrt{1},\sqrt{2},\sqrt{3},\sqrt{4}\)
B \(\sqrt{2},\sqrt{1},\sqrt{3},\sqrt{4}\)
C \(\sqrt{1},\sqrt{3},\sqrt{5},\sqrt{7}\)
D \(\sqrt{4},\sqrt{3},\sqrt{2},\sqrt{1}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{1},\sqrt{2},\sqrt{3},\sqrt{4}\)
Step 1
Concept
The hypotenuses progress as \(\sqrt{1},\sqrt{2},\sqrt{3}\) and so on. In the spiral the number increases by one.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{1},\sqrt{2},\sqrt{3},\sqrt{4}\). The hypotenuses progress as \(\sqrt{1},\sqrt{2},\sqrt{3}\) and so on. In the spiral the number increases by one.
Step 3
Exam Tip
कर्ण क्रम से \(\sqrt{1},\sqrt{2},\sqrt{3}\) आगे बढ़ते हैं। सर्पिल में संख्या एक-एक बढ़ती है।
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यदि पिछले कर्ण की लंबाई \(\sqrt{8}\) है, तो (1) इकाई लंब जोड़ने पर नया कर्ण क्या होगा?
If the previous hypotenuse is \(\sqrt{8}\), what is the new hypotenuse after adding a (1) unit perpendicular?
#square-root-spiral
#next-root
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A \(\sqrt{7}\)
B \(\sqrt{8}\)
C \(\sqrt{9}\)
D \(\sqrt{10}\)
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Correct Answer
C. \(\sqrt{9}\)
Step 1
Concept
The new hypotenuse is \(\sqrt{8+1}=\sqrt{9}\). This is the simple rule of the square root spiral.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{9}\). The new hypotenuse is \(\sqrt{8+1}=\sqrt{9}\). This is the simple rule of the square root spiral.
Step 3
Exam Tip
नया कर्ण \(\sqrt{8+1}=\sqrt{9}\) होगा। यह वर्गमूल सर्पिल का सरल नियम है।
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वर्गमूल सर्पिल में \(\sqrt{12}\) बनाने के लिए पिछला कर्ण कौन-सा होना चाहिए?
To make \(\sqrt{12}\) in a square root spiral, what should be the previous hypotenuse?
#square-root-spiral
#previous-root
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A \(\sqrt{10}\)
B \(\sqrt{11}\)
C \(\sqrt{12}\)
D \(\sqrt{13}\)
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Correct Answer
B. \(\sqrt{11}\)
Step 1
Concept
Adding a (1) unit perpendicular to \(\sqrt{11}\) gives \(\sqrt{12}\). The previous hypotenuse has one less number.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{11}\). Adding a (1) unit perpendicular to \(\sqrt{11}\) gives \(\sqrt{12}\). The previous hypotenuse has one less number.
Step 3
Exam Tip
\(\sqrt{11}\) पर (1) इकाई लंब जोड़ने से \(\sqrt{12}\) बनता है। पिछला कर्ण एक कम संख्या का होता है।
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वर्गमूल सर्पिल में \(\sqrt{16}\) किस संख्या के बराबर है?
In a square root spiral, \(\sqrt{16}\) is equal to which number?
#square-root-spiral
#perfect-square
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A (2)
B (3)
C (4)
D (8)
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Step 1
Concept
\(\sqrt{16}=4\). Square roots of perfect squares give whole numbers.
Step 2
Why this answer is correct
The correct answer is C. (4). \(\sqrt{16}=4\). Square roots of perfect squares give whole numbers.
Step 3
Exam Tip
\(\sqrt{16}=4\) होता है। पूर्ण वर्गों के वर्गमूल पूर्ण संख्याएँ देते हैं।
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वर्गमूल सर्पिल में \(\sqrt{5}\) किस प्रकार की संख्या है?
In a square root spiral, what type of number is \(\sqrt{5}\)?
#square-root-spiral
#irrational-number
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A पूर्ण संख्या / Whole number
B शून्य / Zero
C अपरिमेय संख्या / Irrational number
D ऋणात्मक पूर्णांक / Negative integer
Explanation opens after your attempt
Correct Answer
C. अपरिमेय संख्या / Irrational number
Step 1
Concept
\(\sqrt{5}\) is not the square root of a perfect square. Therefore it is irrational.
Step 2
Why this answer is correct
The correct answer is C. अपरिमेय संख्या / Irrational number. \(\sqrt{5}\) is not the square root of a perfect square. Therefore it is irrational.
Step 3
Exam Tip
\(\sqrt{5}\) पूर्ण वर्ग का वर्गमूल नहीं है। इसलिए यह अपरिमेय संख्या है।
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वर्गमूल सर्पिल में समकोण बनाने के लिए कितने डिग्री का कोण चाहिए?
What angle is needed to make a right angle in a square root spiral?
#square-root-spiral
#right-angle
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A \(30^\circ\)
B \(45^\circ\)
C \(60^\circ\)
D \(90^\circ\)
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Correct Answer
D. \(90^\circ\)
Step 1
Concept
A right angle is \(90^\circ\). Pythagoras theorem applies to this angle.
Step 2
Why this answer is correct
The correct answer is D. \(90^\circ\). A right angle is \(90^\circ\). Pythagoras theorem applies to this angle.
Step 3
Exam Tip
समकोण \(90^\circ\) का होता है। इसी पर पाइथागोरस प्रमेय लागू होता है।
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\(\sqrt{n}\) कर्ण पर (1) इकाई लंब जोड़ने पर नया कर्ण सामान्यतः क्या होगा?
If a (1) unit perpendicular is added to the hypotenuse \(\sqrt{n}\), what will the new hypotenuse generally be?
#square-root-spiral
#general-rule
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A \(\sqrt{n-1}\)
B \(\sqrt{n+1}\)
C \(\sqrt{2n}\)
D \(\sqrt{n^2}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{n+1}\)
Step 1
Concept
By Pythagoras the new hypotenuse is (\sqrt{\(\sqrt{n}\)2 +12 }=\sqrt{n+1}). This is the general rule.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{n+1}\). By Pythagoras the new hypotenuse is (\sqrt{\(\sqrt{n}\)2 +12 }=\sqrt{n+1}). This is the general rule.
Step 3
Exam Tip
पाइथागोरस से नया कर्ण (\sqrt{\(\sqrt{n}\)2 +12 }=\sqrt{n+1}) होता है। यही सामान्य नियम है।
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वर्गमूल सर्पिल में \(\sqrt{20}\) के बाद अगला कर्ण कौन-सा होगा?
In a square root spiral, which hypotenuse comes after \(\sqrt{20}\)?
#square-root-spiral
#next-root
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A \(\sqrt{19}\)
B \(\sqrt{20}\)
C \(\sqrt{21}\)
D \(\sqrt{40}\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{21}\)
Step 1
Concept
At every step the next square root appears. Therefore \(\sqrt{21}\) comes after \(\sqrt{20}\).
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{21}\). At every step the next square root appears. Therefore \(\sqrt{21}\) comes after \(\sqrt{20}\).
Step 3
Exam Tip
हर चरण में अगला वर्गमूल आता है। इसलिए \(\sqrt{20}\) के बाद \(\sqrt{21}\) होगा।
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वर्गमूल सर्पिल से मिली लंबाई को संख्या रेखा पर रखने के लिए कौन-सा उपकरण उपयोगी है?
Which tool is useful to place the length obtained from a square root spiral on the number line?
#square-root-spiral
#compass
#number-line
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A कंपास / Compass
B घड़ी / Clock
C तराजू / Balance
D कैलकुलेटर / Calculator
Explanation opens after your attempt
Correct Answer
A. कंपास / Compass
Step 1
Concept
A compass transfers the hypotenuse length to the number line. This gives the correct point.
Step 2
Why this answer is correct
The correct answer is A. कंपास / Compass. A compass transfers the hypotenuse length to the number line. This gives the correct point.
Step 3
Exam Tip
कंपास से कर्ण की लंबाई संख्या रेखा पर स्थानांतरित की जाती है। इससे सही बिंदु मिलता है।
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वर्गमूल सर्पिल में \(\sqrt{2}\) संख्या रेखा पर किन संख्याओं के बीच होता है?
In a square root spiral, \(\sqrt{2}\) lies between which numbers on the number line?
#square-root-spiral
#number-line
#sqrt2
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A (0) और (1) / (0) and (1)
B (1) और (2) / (1) and (2)
C (2) और (3) / (2) and (3)
D (3) और (4) / (3) and (4)
Explanation opens after your attempt
Correct Answer
B. (1) और (2) / (1) and (2)
Step 1
Concept
Because \(1^2<2<2^2\). Therefore \(1<\sqrt{2}<2\).
Step 2
Why this answer is correct
The correct answer is B. (1) और (2) / (1) and (2). Because \(1^2<2<2^2\). Therefore \(1<\sqrt{2}<2\).
Step 3
Exam Tip
क्योंकि \(1^2<2<2^2\) है। इसलिए \(1<\sqrt{2}<2\) होता है।
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\(\sqrt{10}\) संख्या रेखा पर किन पूर्ण संख्याओं के बीच होगा?
Between which whole numbers will \(\sqrt{10}\) lie on the number line?
#square-root-spiral
#number-line
#interval
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A (1) और (2) / (1) and (2)
B (2) और (3) / (2) and (3)
C (3) और (4) / (3) and (4)
D (4) और (5) / (4) and (5)
Explanation opens after your attempt
Correct Answer
C. (3) और (4) / (3) and (4)
Step 1
Concept
Because \(3^2<10<4^2\). Therefore \(3<\sqrt{10}<4\).
Step 2
Why this answer is correct
The correct answer is C. (3) और (4) / (3) and (4). Because \(3^2<10<4^2\). Therefore \(3<\sqrt{10}<4\).
Step 3
Exam Tip
क्योंकि \(3^2<10<4^2\) है। इसलिए \(3<\sqrt{10}<4\) होता है।
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वर्गमूल सर्पिल में \(\sqrt{2}+1=\sqrt{3}\) मानना क्यों गलत है?
Why is it wrong to assume \(\sqrt{2}+1=\sqrt{3}\) in a square root spiral?
#square-root-spiral
#common-error
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A क्योंकि नया कर्ण वर्गों के योग से बनता है / Because the new hypotenuse is formed by the sum of squares
B क्योंकि \(\sqrt{2}=1\) होता है / Because \(\sqrt{2}=1\)
C क्योंकि (1=0) होता है / Because (1=0)
D क्योंकि \(\sqrt{3}=2\) होता है / Because \(\sqrt{3}=2\)
Explanation opens after your attempt
Correct Answer
A. क्योंकि नया कर्ण वर्गों के योग से बनता है / Because the new hypotenuse is formed by the sum of squares
Step 1
Concept
In a square root spiral the addition is of squares, not direct lengths. The correct reasoning is Pythagoras theorem.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि नया कर्ण वर्गों के योग से बनता है / Because the new hypotenuse is formed by the sum of squares. In a square root spiral the addition is of squares, not direct lengths. The correct reasoning is Pythagoras theorem.
Step 3
Exam Tip
वर्गमूल सर्पिल में जोड़ सीधे लंबाइयों का नहीं, वर्गों का होता है। सही तर्क पाइथागोरस प्रमेय है।
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वर्गमूल सर्पिल में बनने वाली आकृति कैसी दिखाई देती है?
What does the figure formed in a square root spiral look like?
#square-root-spiral
#shape
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A सर्पिल जैसी / Like a spiral
B सीधी रेखा जैसी / Like a straight line
C वर्ग जैसी / Like a square
D केवल वृत्त जैसी / Like only a circle
Explanation opens after your attempt
Correct Answer
A. सर्पिल जैसी / Like a spiral
Step 1
Concept
Successive right triangles are formed while changing direction. Therefore the figure looks like a spiral.
Step 2
Why this answer is correct
The correct answer is A. सर्पिल जैसी / Like a spiral. Successive right triangles are formed while changing direction. Therefore the figure looks like a spiral.
Step 3
Exam Tip
लगातार समकोण त्रिभुज दिशा बदलते हुए बनते हैं। इसलिए आकृति सर्पिल जैसी दिखती है।
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वर्गमूल सर्पिल का मुख्य उद्देश्य क्या है?
What is the main purpose of a square root spiral?
#square-root-spiral
#purpose
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A वर्गमूल लंबाइयों का ज्यामितीय निर्माण करना / To geometrically construct square root lengths
B केवल गुणा करना / To only multiply
C केवल प्रतिशत निकालना / To only find percentages
D हर को शून्य बनाना / To make denominator zero
Explanation opens after your attempt
Correct Answer
A. वर्गमूल लंबाइयों का ज्यामितीय निर्माण करना / To geometrically construct square root lengths
Step 1
Concept
A square root spiral constructs lengths like \(\sqrt{2},\sqrt{3},\sqrt{5}\). It is useful on the number line.
Step 2
Why this answer is correct
The correct answer is A. वर्गमूल लंबाइयों का ज्यामितीय निर्माण करना / To geometrically construct square root lengths. A square root spiral constructs lengths like \(\sqrt{2},\sqrt{3},\sqrt{5}\). It is useful on the number line.
Step 3
Exam Tip
वर्गमूल सर्पिल \(\sqrt{2},\sqrt{3},\sqrt{5}\) जैसी लंबाइयाँ बनाता है। यह संख्या रेखा पर उपयोगी है।
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वर्गमूल सर्पिल में \(\sqrt{18}\) बनाने से पहले कौन-सा कर्ण बन चुका होगा?
Before constructing \(\sqrt{18}\) in a square root spiral, which hypotenuse will already be constructed?
#square-root-spiral
#previous-root
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A \(\sqrt{16}\)
B \(\sqrt{17}\)
C \(\sqrt{18}\)
D \(\sqrt{19}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{17}\)
Step 1
Concept
Adding a (1) unit perpendicular to \(\sqrt{17}\) gives \(\sqrt{18}\). The previous hypotenuse has one less number.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{17}\). Adding a (1) unit perpendicular to \(\sqrt{17}\) gives \(\sqrt{18}\). The previous hypotenuse has one less number.
Step 3
Exam Tip
\(\sqrt{17}\) में (1) इकाई लंब जोड़कर \(\sqrt{18}\) मिलता है। पिछले कर्ण की संख्या एक कम होती है।
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वर्गमूल सर्पिल में \(\sqrt{30}\) के बाद कौन-सा कर्ण आएगा?
In a square root spiral, which hypotenuse comes after \(\sqrt{30}\)?
#square-root-spiral
#next-root
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A \(\sqrt{29}\)
B \(\sqrt{30}\)
C \(\sqrt{31}\)
D \(\sqrt{60}\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{31}\)
Step 1
Concept
In the next step the number increases by (1). Therefore \(\sqrt{31}\) comes after \(\sqrt{30}\).
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{31}\). In the next step the number increases by (1). Therefore \(\sqrt{31}\) comes after \(\sqrt{30}\).
Step 3
Exam Tip
अगले चरण में संख्या (1) बढ़ती है। इसलिए \(\sqrt{30}\) के बाद \(\sqrt{31}\) आता है।
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वर्गमूल सर्पिल में \(\sqrt{36}\) किसके बराबर होगा?
In a square root spiral, \(\sqrt{36}\) will be equal to what?
#square-root-spiral
#perfect-square
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A (5)
B (6)
C (7)
D (8)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{36}=6\). The number (36) is a perfect square.
Step 2
Why this answer is correct
The correct answer is B. (6). \(\sqrt{36}=6\). The number (36) is a perfect square.
Step 3
Exam Tip
\(\sqrt{36}=6\) होता है। (36) पूर्ण वर्ग है।
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\(\sqrt{26}\) संख्या रेखा पर किन पूर्ण संख्याओं के बीच होगा?
Between which whole numbers will \(\sqrt{26}\) lie on the number line?
#square-root-spiral
#number-line
#interval
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A (4) और (5) / (4) and (5)
B (5) और (6) / (5) and (6)
C (6) और (7) / (6) and (7)
D (7) और (8) / (7) and (8)
Explanation opens after your attempt
Correct Answer
B. (5) और (6) / (5) and (6)
Step 1
Concept
Because \(5^2<26<6^2\). Therefore \(5<\sqrt{26}<6\).
Step 2
Why this answer is correct
The correct answer is B. (5) और (6) / (5) and (6). Because \(5^2<26<6^2\). Therefore \(5<\sqrt{26}<6\).
Step 3
Exam Tip
क्योंकि \(5^2<26<6^2\) है। इसलिए \(5<\sqrt{26}<6\) होता है।
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वर्गमूल सर्पिल में (1) इकाई लंब को \(90^\circ\) पर खींचने का कारण क्या है?
Why is the (1) unit perpendicular drawn at \(90^\circ\) in a square root spiral?
#square-root-spiral
#right-angle
#construction
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A समकोण त्रिभुज बनाने के लिए / To make a right triangle
B वृत्त मिटाने के लिए / To erase a circle
C रेखा को आधा करने के लिए / To halve the line
D कर्ण को शून्य करने के लिए / To make the hypotenuse zero
Explanation opens after your attempt
Correct Answer
A. समकोण त्रिभुज बनाने के लिए / To make a right triangle
Step 1
Concept
A \(90^\circ\) angle makes a right triangle. Then Pythagoras theorem gives the new hypotenuse.
Step 2
Why this answer is correct
The correct answer is A. समकोण त्रिभुज बनाने के लिए / To make a right triangle. A \(90^\circ\) angle makes a right triangle. Then Pythagoras theorem gives the new hypotenuse.
Step 3
Exam Tip
\(90^\circ\) से समकोण त्रिभुज बनता है। फिर पाइथागोरस प्रमेय से नया कर्ण मिलता है।
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वर्गमूल सर्पिल में \(\sqrt{50}\) बनाने के लिए कौन-सा पिछला कर्ण उपयोग होगा?
Which previous hypotenuse is used to construct \(\sqrt{50}\) in a square root spiral?
#square-root-spiral
#previous-root
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A \(\sqrt{48}\)
B \(\sqrt{49}\)
C \(\sqrt{50}\)
D \(\sqrt{51}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{49}\)
Step 1
Concept
Adding a (1) unit perpendicular to \(\sqrt{49}\) gives \(\sqrt{50}\). The previous number is (1) less.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{49}\). Adding a (1) unit perpendicular to \(\sqrt{49}\) gives \(\sqrt{50}\). The previous number is (1) less.
Step 3
Exam Tip
\(\sqrt{49}\) में (1) इकाई लंब जोड़ने पर \(\sqrt{50}\) बनता है। पिछली संख्या (1) कम होती है।
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वर्गमूल सर्पिल में \(\sqrt{49}\) की लंबाई क्या होगी?
What is the length of \(\sqrt{49}\) in a square root spiral?
#square-root-spiral
#perfect-square
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A (6)
B (7)
C (8)
D (9)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{49}=7\). The number (49) is a perfect square.
Step 2
Why this answer is correct
The correct answer is B. (7). \(\sqrt{49}=7\). The number (49) is a perfect square.
Step 3
Exam Tip
\(\sqrt{49}=7\) होता है। (49) पूर्ण वर्ग है।
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\(\sqrt{40}\) संख्या रेखा पर किन पूर्ण संख्याओं के बीच होगा?
Between which whole numbers will \(\sqrt{40}\) lie on the number line?
#square-root-spiral
#number-line
#interval
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A (5) और (6) / (5) and (6)
B (6) और (7) / (6) and (7)
C (7) और (8) / (7) and (8)
D (8) और (9) / (8) and (9)
Explanation opens after your attempt
Correct Answer
B. (6) और (7) / (6) and (7)
Step 1
Concept
Because \(6^2<40<7^2\). Therefore \(6<\sqrt{40}<7\).
Step 2
Why this answer is correct
The correct answer is B. (6) और (7) / (6) and (7). Because \(6^2<40<7^2\). Therefore \(6<\sqrt{40}<7\).
Step 3
Exam Tip
क्योंकि \(6^2<40<7^2\) है। इसलिए \(6<\sqrt{40}<7\) होता है।
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वर्गमूल सर्पिल में \(\sqrt{6}\) से नया कर्ण बनाने पर कौन-सा कर्ण मिलेगा?
If a new hypotenuse is made from \(\sqrt{6}\) in a square root spiral, which hypotenuse is obtained?
#square-root-spiral
#next-root
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A \(\sqrt{5}\)
B \(\sqrt{6}\)
C \(\sqrt{7}\)
D \(\sqrt{12}\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{7}\)
Step 1
Concept
Adding a (1) unit perpendicular to \(\sqrt{6}\) forms \(\sqrt{7}\). This follows from Pythagoras theorem.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{7}\). Adding a (1) unit perpendicular to \(\sqrt{6}\) forms \(\sqrt{7}\). This follows from Pythagoras theorem.
Step 3
Exam Tip
\(\sqrt{6}\) में (1) इकाई लंब जोड़ने से \(\sqrt{7}\) बनता है। यह पाइथागोरस प्रमेय से आता है।
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वर्गमूल सर्पिल में \(\sqrt{64}\) किस संख्या के बराबर है?
In a square root spiral, \(\sqrt{64}\) is equal to which number?
#square-root-spiral
#perfect-square
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A (6)
B (7)
C (8)
D (9)
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Step 1
Concept
\(\sqrt{64}=8\). Recognizing perfect squares gives the answer quickly.
Step 2
Why this answer is correct
The correct answer is C. (8). \(\sqrt{64}=8\). Recognizing perfect squares gives the answer quickly.
Step 3
Exam Tip
\(\sqrt{64}=8\) होता है। पूर्ण वर्गों की पहचान से उत्तर जल्दी मिलता है।
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वर्गमूल सर्पिल में \(\sqrt{2}\) किस प्रकार की संख्या है?
In a square root spiral, what type of number is \(\sqrt{2}\)?
#square-root-spiral
#irrational-number
#sqrt2
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A अपरिमेय संख्या / Irrational number
B पूर्ण संख्या / Whole number
C शून्य / Zero
D ऋणात्मक संख्या / Negative number
Explanation opens after your attempt
Correct Answer
A. अपरिमेय संख्या / Irrational number
Step 1
Concept
\(\sqrt{2}\) is not the square root of a perfect square. Therefore it is irrational.
Step 2
Why this answer is correct
The correct answer is A. अपरिमेय संख्या / Irrational number. \(\sqrt{2}\) is not the square root of a perfect square. Therefore it is irrational.
Step 3
Exam Tip
\(\sqrt{2}\) किसी पूर्ण वर्ग का वर्गमूल नहीं है। इसलिए यह अपरिमेय संख्या है।
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वर्गमूल सर्पिल में \(\sqrt{15}\) बनाने से ठीक पहले कौन-सा कर्ण होगा?
Just before constructing \(\sqrt{15}\) in a square root spiral, which hypotenuse will be present?
#square-root-spiral
#previous-root
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A \(\sqrt{13}\)
B \(\sqrt{14}\)
C \(\sqrt{15}\)
D \(\sqrt{16}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{14}\)
Step 1
Concept
With \(\sqrt{14}\) and a (1) unit perpendicular, \(\sqrt{15}\) is formed. The previous hypotenuse has one less number.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{14}\). With \(\sqrt{14}\) and a (1) unit perpendicular, \(\sqrt{15}\) is formed. The previous hypotenuse has one less number.
Step 3
Exam Tip
\(\sqrt{14}\) के साथ (1) इकाई लंब से \(\sqrt{15}\) बनता है। पिछला कर्ण एक कम संख्या का है।
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वर्गमूल सर्पिल में \(\sqrt{24}\) के बाद कौन-सा कर्ण बनेगा?
In a square root spiral, which hypotenuse will be formed after \(\sqrt{24}\)?
#square-root-spiral
#next-root
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A \(\sqrt{23}\)
B \(\sqrt{24}\)
C \(\sqrt{25}\)
D \(\sqrt{26}\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{25}\)
Step 1
Concept
In the next step \(\sqrt{24+1}=\sqrt{25}\) is formed. The number increases by (1).
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{25}\). In the next step \(\sqrt{24+1}=\sqrt{25}\) is formed. The number increases by (1).
Step 3
Exam Tip
अगले चरण में \(\sqrt{24+1}=\sqrt{25}\) बनता है। संख्या क्रम में (1) बढ़ती है।
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वर्गमूल सर्पिल से \(\sqrt{3}\) को संख्या रेखा पर अंकित करने के लिए किस लंबाई को कंपास में लेना होगा?
To mark \(\sqrt{3}\) on the number line from a square root spiral, which length should be taken in the compass?
#square-root-spiral
#number-line
#compass
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A \(\sqrt{2}\) का कर्ण / Hypotenuse of \(\sqrt{2}\)
B \(\sqrt{3}\) का कर्ण / Hypotenuse of \(\sqrt{3}\)
C (3) इकाई रेखा / (3) unit line
D (1) इकाई लंब / (1) unit perpendicular
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{3}\) का कर्ण / Hypotenuse of \(\sqrt{3}\)
Step 1
Concept
The hypotenuse length of the required square root is taken in the compass. So the hypotenuse of \(\sqrt{3}\) is needed.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{3}\) का कर्ण / Hypotenuse of \(\sqrt{3}\). The hypotenuse length of the required square root is taken in the compass. So the hypotenuse of \(\sqrt{3}\) is needed.
Step 3
Exam Tip
जिस वर्गमूल को अंकित करना है, उसी कर्ण की लंबाई कंपास में लेते हैं। इसलिए \(\sqrt{3}\) का कर्ण चाहिए।
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वर्गमूल सर्पिल में \(\sqrt{3}\) किन पूर्ण संख्याओं के बीच स्थित है?
In a square root spiral, \(\sqrt{3}\) lies between which whole numbers?
#square-root-spiral
#number-line
#interval
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A (0) और (1) / (0) and (1)
B (1) और (2) / (1) and (2)
C (2) और (3) / (2) and (3)
D (3) और (4) / (3) and (4)
Explanation opens after your attempt
Correct Answer
B. (1) और (2) / (1) and (2)
Step 1
Concept
Because \(1^2<3<2^2\). Therefore \(\sqrt{3}\) lies between (1) and (2).
Step 2
Why this answer is correct
The correct answer is B. (1) और (2) / (1) and (2). Because \(1^2<3<2^2\). Therefore \(\sqrt{3}\) lies between (1) and (2).
Step 3
Exam Tip
क्योंकि \(1^2<3<2^2\) है। इसलिए \(\sqrt{3}\) (1) और (2) के बीच है।
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वर्गमूल सर्पिल में सेट स्क्वायर का उपयोग किसलिए किया जा सकता है?
For what can a set square be used in a square root spiral?
#square-root-spiral
#tools
#right-angle
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A \(90^\circ\) का कोण बनाने के लिए / To make a \(90^\circ\) angle
B कर्ण मिटाने के लिए / To erase the hypotenuse
C वर्गमूल हटाने के लिए / To remove square roots
D दशमलव निकालने के लिए / To find decimals
Explanation opens after your attempt
Correct Answer
A. \(90^\circ\) का कोण बनाने के लिए / To make a \(90^\circ\) angle
Step 1
Concept
A set square helps make a right angle easily. A correct right angle gives the correct hypotenuse.
Step 2
Why this answer is correct
The correct answer is A. \(90^\circ\) का कोण बनाने के लिए / To make a \(90^\circ\) angle. A set square helps make a right angle easily. A correct right angle gives the correct hypotenuse.
Step 3
Exam Tip
सेट स्क्वायर से समकोण आसानी से बनता है। सही समकोण से सही कर्ण मिलता है।
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वर्गमूल सर्पिल में \(\sqrt{72}\) किस दो पूर्ण संख्याओं के बीच होगा?
In a square root spiral, \(\sqrt{72}\) lies between which two whole numbers?
#square-root-spiral
#number-line
#interval
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A (6) और (7) / (6) and (7)
B (7) और (8) / (7) and (8)
C (8) और (9) / (8) and (9)
D (9) और (10) / (9) and (10)
Explanation opens after your attempt
Correct Answer
C. (8) और (9) / (8) and (9)
Step 1
Concept
Because \(8^2<72<9^2\). Therefore \(8<\sqrt{72}<9\).
Step 2
Why this answer is correct
The correct answer is C. (8) और (9) / (8) and (9). Because \(8^2<72<9^2\). Therefore \(8<\sqrt{72}<9\).
Step 3
Exam Tip
क्योंकि \(8^2<72<9^2\) है। इसलिए \(8<\sqrt{72}<9\) होता है।
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वर्गमूल सर्पिल में \(\sqrt{80}\) बनाने के लिए पिछला कर्ण कौन-सा होगा?
To construct \(\sqrt{80}\) in a square root spiral, which previous hypotenuse will be used?
#square-root-spiral
#previous-root
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A \(\sqrt{78}\)
B \(\sqrt{79}\)
C \(\sqrt{80}\)
D \(\sqrt{81}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{79}\)
Step 1
Concept
Adding a (1) unit perpendicular to \(\sqrt{79}\) gives \(\sqrt{80}\). The previous hypotenuse has one less number.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{79}\). Adding a (1) unit perpendicular to \(\sqrt{79}\) gives \(\sqrt{80}\). The previous hypotenuse has one less number.
Step 3
Exam Tip
\(\sqrt{79}\) पर (1) इकाई लंब जोड़ने से \(\sqrt{80}\) बनता है। पिछला कर्ण एक कम संख्या का होता है।
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वर्गमूल सर्पिल में \(\sqrt{81}\) की लंबाई क्या होगी?
What is the length of \(\sqrt{81}\) in a square root spiral?
#square-root-spiral
#perfect-square
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A (7)
B (8)
C (9)
D (10)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{81}=9\). The number (81) is a perfect square.
Step 2
Why this answer is correct
The correct answer is C. (9). \(\sqrt{81}=9\). The number (81) is a perfect square.
Step 3
Exam Tip
\(\sqrt{81}=9\) होता है। (81) पूर्ण वर्ग है।
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वर्गमूल सर्पिल में \(\sqrt{2}\) बनाने के बाद कौन-सा निर्माण करना होता है?
After constructing \(\sqrt{2}\) in a square root spiral, what construction is done next?
#square-root-spiral
#construction-step
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A \(\sqrt{2}\) के सिरे पर (1) इकाई लंब बनाना / Draw a (1) unit perpendicular at the end of \(\sqrt{2}\)
B रेखाखंड को मिटाना / Erase the segment
C हर को शून्य करना / Make the denominator zero
D त्रिभुज को वर्ग बनाना / Convert the triangle into a square
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{2}\) के सिरे पर (1) इकाई लंब बनाना / Draw a (1) unit perpendicular at the end of \(\sqrt{2}\)
Step 1
Concept
A (1) unit perpendicular is drawn at the end of \(\sqrt{2}\) to make the next triangle. This gives \(\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}\) के सिरे पर (1) इकाई लंब बनाना / Draw a (1) unit perpendicular at the end of \(\sqrt{2}\). A (1) unit perpendicular is drawn at the end of \(\sqrt{2}\) to make the next triangle. This gives \(\sqrt{3}\).
Step 3
Exam Tip
\(\sqrt{2}\) के सिरे पर (1) इकाई लंब बनाकर अगला त्रिभुज बनाया जाता है। इससे \(\sqrt{3}\) मिलता है।
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वर्गमूल सर्पिल में \(\sqrt{4}\) के बाद बनने वाला \(\sqrt{5}\) किस प्रकार की संख्या है?
In a square root spiral, what type of number is \(\sqrt{5}\) formed after \(\sqrt{4}\)?
#square-root-spiral
#irrational-number
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A अपरिमेय संख्या / Irrational number
B पूर्ण संख्या / Whole number
C शून्य / Zero
D ऋणात्मक संख्या / Negative number
Explanation opens after your attempt
Correct Answer
A. अपरिमेय संख्या / Irrational number
Step 1
Concept
\(\sqrt{5}\) is not the square root of a perfect square. Therefore it is irrational.
Step 2
Why this answer is correct
The correct answer is A. अपरिमेय संख्या / Irrational number. \(\sqrt{5}\) is not the square root of a perfect square. Therefore it is irrational.
Step 3
Exam Tip
\(\sqrt{5}\) पूर्ण वर्ग का वर्गमूल नहीं है। इसलिए यह अपरिमेय है।
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वर्गमूल सर्पिल में \(\sqrt{100}\) किस संख्या के बराबर होगा?
In a square root spiral, \(\sqrt{100}\) will be equal to which number?
#square-root-spiral
#perfect-square
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A (9)
B (10)
C (11)
D (100)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{100}=10\). Square roots of perfect squares give whole numbers.
Step 2
Why this answer is correct
The correct answer is B. (10). \(\sqrt{100}=10\). Square roots of perfect squares give whole numbers.
Step 3
Exam Tip
\(\sqrt{100}=10\) होता है। पूर्ण वर्गों के वर्गमूल पूर्ण संख्या देते हैं।
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वर्गमूल सर्पिल में \(\sqrt{101}\) किस दो पूर्ण संख्याओं के बीच होगा?
In a square root spiral, \(\sqrt{101}\) lies between which two whole numbers?
#square-root-spiral
#number-line
#interval
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A (9) और (10) / (9) and (10)
B (10) और (11) / (10) and (11)
C (11) और (12) / (11) and (12)
D (12) और (13) / (12) and (13)
Explanation opens after your attempt
Correct Answer
B. (10) और (11) / (10) and (11)
Step 1
Concept
Because \(10^2<101<11^2\). Therefore \(10<\sqrt{101}<11\).
Step 2
Why this answer is correct
The correct answer is B. (10) और (11) / (10) and (11). Because \(10^2<101<11^2\). Therefore \(10<\sqrt{101}<11\).
Step 3
Exam Tip
क्योंकि \(10^2<101<11^2\) है। इसलिए \(10<\sqrt{101}<11\) होता है।
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वर्गमूल सर्पिल में कौन-सा कथन \(\sqrt{n}\) से \(\sqrt{n+1}\) बनने को सही समझाता है?
Which statement correctly explains how \(\sqrt{n}\) becomes \(\sqrt{n+1}\) in a square root spiral?
#square-root-spiral
#general-rule
#pythagoras
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A पिछले कर्ण और (1) इकाई लंब के वर्गों का योग लिया जाता है / The sum of squares of previous hypotenuse and (1) unit perpendicular is taken
B पिछले कर्ण में सीधे (1) जोड़ दिया जाता है / (1) is directly added to the previous hypotenuse
C पिछले कर्ण को (2) से गुणा किया जाता है / The previous hypotenuse is multiplied by (2)
D पिछले कर्ण को (1) से घटाया जाता है / The previous hypotenuse is decreased by (1)
Explanation opens after your attempt
Correct Answer
A. पिछले कर्ण और (1) इकाई लंब के वर्गों का योग लिया जाता है / The sum of squares of previous hypotenuse and (1) unit perpendicular is taken
Step 1
Concept
Pythagoras theorem uses the sum of squares. Therefore the new hypotenuse becomes \(\sqrt{n+1}\).
Step 2
Why this answer is correct
The correct answer is A. पिछले कर्ण और (1) इकाई लंब के वर्गों का योग लिया जाता है / The sum of squares of previous hypotenuse and (1) unit perpendicular is taken. Pythagoras theorem uses the sum of squares. Therefore the new hypotenuse becomes \(\sqrt{n+1}\).
Step 3
Exam Tip
पाइथागोरस प्रमेय में वर्गों का योग आता है। इसलिए नया कर्ण \(\sqrt{n+1}\) बनता है।
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वर्गमूल सर्पिल में \(\sqrt{2}\) की लंबाई संख्या रेखा पर अंकित करते समय चाप कहाँ से खींचा जाता है?
While marking \(\sqrt{2}\) on the number line from a square root spiral, from where is the arc drawn?
#square-root-spiral
#number-line
#arc
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A मूल बिंदु से / From the origin
B किसी भी ऋणात्मक बिंदु से / From any negative point
C कर्ण के मध्यबिंदु से / From the midpoint of the hypotenuse
D त्रिभुज के अंदर से / From inside the triangle
Explanation opens after your attempt
Correct Answer
A. मूल बिंदु से / From the origin
Step 1
Concept
The compass point is kept at the origin and an arc is drawn. This locates \(\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. मूल बिंदु से / From the origin. The compass point is kept at the origin and an arc is drawn. This locates \(\sqrt{2}\).
Step 3
Exam Tip
कंपास की नोक मूल बिंदु पर रखकर चाप खींचते हैं। इससे \(\sqrt{2}\) का स्थान मिलता है।
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वर्गमूल सर्पिल का मुख्य विचार एक पंक्ति में कौन-सा है?
What is the main idea of a square root spiral in one line?
#square-root-spiral
#main-idea
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A समकोण त्रिभुजों से क्रमिक वर्गमूल लंबाइयाँ बनाना / Making successive square root lengths using right triangles
B केवल वृत्त बनाना / Making only circles
C केवल दशमलव मान याद करना / Memorizing only decimal values
D केवल ऋणात्मक संख्याएँ बनाना / Making only negative numbers
Explanation opens after your attempt
Correct Answer
A. समकोण त्रिभुजों से क्रमिक वर्गमूल लंबाइयाँ बनाना / Making successive square root lengths using right triangles
Step 1
Concept
Each new right triangle gives the next square root. This is the basic idea of a square root spiral.
Step 2
Why this answer is correct
The correct answer is A. समकोण त्रिभुजों से क्रमिक वर्गमूल लंबाइयाँ बनाना / Making successive square root lengths using right triangles. Each new right triangle gives the next square root. This is the basic idea of a square root spiral.
Step 3
Exam Tip
हर नया समकोण त्रिभुज अगला वर्गमूल देता है। यही वर्गमूल सर्पिल का मूल विचार है।
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वर्गमूल सर्पिल में \(\sqrt{121}\) की लंबाई किस संख्या के बराबर होगी?
In a square root spiral, the length \(\sqrt{121}\) will be equal to which number?
#square-root-spiral
#perfect-square
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A (10)
B (11)
C (12)
D (121)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{121}=11\). Since (121) is a perfect square, its square root is a whole number.
Step 2
Why this answer is correct
The correct answer is B. (11). \(\sqrt{121}=11\). Since (121) is a perfect square, its square root is a whole number.
Step 3
Exam Tip
\(\sqrt{121}=11\) होता है। (121) पूर्ण वर्ग है, इसलिए उसका वर्गमूल पूर्ण संख्या है।
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वर्गमूल सर्पिल में \(\sqrt{57}\) बनाने के लिए किस पिछले कर्ण पर (1) इकाई लंब बनाई जाएगी?
To construct \(\sqrt{57}\) in a square root spiral, a (1) unit perpendicular will be drawn on which previous hypotenuse?
#square-root-spiral
#previous-root
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A \(\sqrt{55}\)
B \(\sqrt{56}\)
C \(\sqrt{57}\)
D \(\sqrt{58}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{56}\)
Step 1
Concept
Adding a (1) unit perpendicular to \(\sqrt{56}\) forms \(\sqrt{57}\). The previous hypotenuse number is always (1) less.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{56}\). Adding a (1) unit perpendicular to \(\sqrt{56}\) forms \(\sqrt{57}\). The previous hypotenuse number is always (1) less.
Step 3
Exam Tip
\(\sqrt{56}\) के साथ (1) इकाई लंब जोड़ने पर \(\sqrt{57}\) बनता है। पिछले कर्ण की संख्या हमेशा (1) कम होती है।
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वर्गमूल सर्पिल में \(\sqrt{43}\) के बाद अगला कर्ण कौन-सा बनेगा?
In a square root spiral, which hypotenuse will be formed after \(\sqrt{43}\)?
#square-root-spiral
#next-root
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A \(\sqrt{42}\)
B \(\sqrt{43}\)
C \(\sqrt{44}\)
D \(\sqrt{86}\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{44}\)
Step 1
Concept
At each new step, a (1) unit perpendicular is added. Therefore \(\sqrt{44}\) will be formed after \(\sqrt{43}\).
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{44}\). At each new step, a (1) unit perpendicular is added. Therefore \(\sqrt{44}\) will be formed after \(\sqrt{43}\).
Step 3
Exam Tip
हर नए चरण में (1) इकाई लंब जुड़ती है। इसलिए \(\sqrt{43}\) के बाद \(\sqrt{44}\) बनेगा।
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वर्गमूल सर्पिल बनाते समय (1) इकाई की दूरी सही मापने के लिए कौन-सा उपकरण उपयोगी है?
Which tool is useful for measuring the (1) unit distance correctly while making a square root spiral?
#square-root-spiral
#tools
#measurement
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A मापक पट्टी / Ruler
B घड़ी / Clock
C कैलकुलेटर / Calculator
D तराजू / Balance
Explanation opens after your attempt
Correct Answer
A. मापक पट्टी / Ruler
Step 1
Concept
A ruler is useful for taking the (1) unit length correctly. Wrong measurement can make the spiral lengths incorrect.
Step 2
Why this answer is correct
The correct answer is A. मापक पट्टी / Ruler. A ruler is useful for taking the (1) unit length correctly. Wrong measurement can make the spiral lengths incorrect.
Step 3
Exam Tip
(1) इकाई की लंबाई सही लेने के लिए मापक पट्टी उपयोगी है। गलत माप से सर्पिल की लंबाइयाँ गलत हो सकती हैं।
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