वर्गमूल सर्पिल बनाने में पहला रेखाखंड सामान्यतः कितनी लंबाई का लिया जाता है?
In making a square root spiral, what length is usually taken for the first line segment?
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A (1) इकाई / (1) unit
B (2) इकाई / (2) units
C (3) इकाई / (3) units
D (4) इकाई / (4) units
Explanation opens after your attempt
Correct Answer
A. (1) इकाई / (1) unit
Step 1
Concept
The first line segment is taken as (1) unit. Right triangles are built from it onward.
Step 2
Why this answer is correct
The correct answer is A. (1) इकाई / (1) unit. The first line segment is taken as (1) unit. Right triangles are built from it onward.
Step 3
Exam Tip
पहला रेखाखंड (1) इकाई लिया जाता है। इसी से आगे समकोण त्रिभुज बनते हैं।
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वर्गमूल सर्पिल में (1) इकाई आधार और (1) इकाई लंब से बने पहले समकोण त्रिभुज का कर्ण कितना होता है?
In a square root spiral, what is the hypotenuse of the first right triangle made with base (1) unit and perpendicular (1) unit?
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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 \(\sqrt{1^2+1^2}=\sqrt{2}\). Remember the first hypotenuse.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{2}\). By Pythagoras theorem, the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). Remember the first hypotenuse.
Step 3
Exam Tip
पाइथागोरस प्रमेय से कर्ण \(\sqrt{1^2+1^2}=\sqrt{2}\) होता है। पहला कर्ण याद रखें।
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\(\sqrt{2}\) लंबाई वाले कर्ण के सिरे पर (1) इकाई लंब खींचकर अगला कर्ण कितना बनेगा?
If a perpendicular of (1) unit is drawn at the end of the hypotenuse of length \(\sqrt{2}\), what will be the next hypotenuse?
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A \(\sqrt{2}\)
B \(\sqrt{3}\)
C \(\sqrt{4}\)
D \(\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{3}\)
Step 1
Concept
The new hypotenuse is (\sqrt{\(\sqrt{2}\)2 +12 }=\sqrt{3}). Each time a (1) unit perpendicular is added.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{3}\). The new hypotenuse is (\sqrt{\(\sqrt{2}\)2 +12 }=\sqrt{3}). Each time a (1) unit perpendicular is added.
Step 3
Exam Tip
नया कर्ण (\sqrt{\(\sqrt{2}\)2 +12 }=\sqrt{3}) होगा। हर बार (1) इकाई लंब जुड़ता है।
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वर्गमूल सर्पिल में \(\sqrt{3}\) कर्ण के बाद (1) इकाई लंब जोड़ने पर कौन-सी लंबाई मिलती है?
In a square root spiral, after the hypotenuse \(\sqrt{3}\), which length is obtained by adding a (1) unit perpendicular?
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A \(\sqrt{4}\)
B \(\sqrt{5}\)
C \(\sqrt{6}\)
D \(\sqrt{2}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{4}\)
Step 1
Concept
With \(\sqrt{3}\) and (1) unit, the new hypotenuse is \(\sqrt{3+1}=\sqrt{4}\). The next square root appears in order.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{4}\). With \(\sqrt{3}\) and (1) unit, the new hypotenuse is \(\sqrt{3+1}=\sqrt{4}\). The next square root appears in order.
Step 3
Exam Tip
\(\sqrt{3}\) और (1) इकाई से नया कर्ण \(\sqrt{3+1}=\sqrt{4}\) बनता है। क्रम में अगला वर्गमूल आता है।
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वर्गमूल सर्पिल का प्रत्येक नया त्रिभुज किस प्रकार का होता है?
Each new triangle in a square root spiral is of which type?
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A समकोण त्रिभुज / Right triangle
B समबाहु त्रिभुज / Equilateral triangle
C विषमबाहु नहीं / Not scalene
D वृत्त / Circle
Explanation opens after your attempt
Correct Answer
A. समकोण त्रिभुज / Right triangle
Step 1
Concept
At every step, a \(90^\circ\) angle is made. Therefore each new triangle is a right triangle.
Step 2
Why this answer is correct
The correct answer is A. समकोण त्रिभुज / Right triangle. At every step, a \(90^\circ\) angle is made. Therefore each new triangle is a right triangle.
Step 3
Exam Tip
हर चरण में \(90^\circ\) का कोण बनाया जाता है। इसलिए प्रत्येक नया त्रिभुज समकोण त्रिभुज होता है।
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वर्गमूल सर्पिल में नए त्रिभुज की नई लंब भुजा सामान्यतः कितनी रखी जाती है?
In a square root spiral, how long is the new perpendicular side usually kept in each new triangle?
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A \(\sqrt{2}\) इकाई / \(\sqrt{2}\) units
B (2) इकाई / (2) units
C (1) इकाई / (1) unit
D \(\sqrt{3}\) इकाई / \(\sqrt{3}\) units
Explanation opens after your attempt
Correct Answer
C. (1) इकाई / (1) unit
Step 1
Concept
In each new triangle, the perpendicular side is taken as (1) unit. This forms the next square root.
Step 2
Why this answer is correct
The correct answer is C. (1) इकाई / (1) unit. In each new triangle, the perpendicular side is taken as (1) unit. This forms the next square root.
Step 3
Exam Tip
हर नए त्रिभुज में लंब भुजा (1) इकाई ली जाती है। इसी से अगला वर्गमूल बनता है।
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वर्गमूल सर्पिल में कर्ण की लंबाई ज्ञात करने के लिए मुख्य रूप से कौन-सा प्रमेय प्रयोग होता है?
Which theorem is mainly used to find the hypotenuse in a square root spiral?
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A थेल्स प्रमेय / Thales theorem
B पाइथागोरस प्रमेय / Pythagoras theorem
C मध्यबिंदु प्रमेय / Midpoint theorem
D शेषफल प्रमेय / Remainder theorem
Explanation opens after your attempt
Correct Answer
B. पाइथागोरस प्रमेय / Pythagoras theorem
Step 1
Concept
Pythagoras theorem is used to find the hypotenuse. Recognizing the right triangle is important.
Step 2
Why this answer is correct
The correct answer is B. पाइथागोरस प्रमेय / Pythagoras theorem. Pythagoras theorem is used to find the hypotenuse. Recognizing the right triangle is important.
Step 3
Exam Tip
कर्ण ज्ञात करने के लिए पाइथागोरस प्रमेय लगाया जाता है। समकोण त्रिभुज पहचानना जरूरी है।
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यदि किसी चरण पर कर्ण \(\sqrt{5}\) है और (1) इकाई लंब जोड़ी जाती है, तो अगला कर्ण क्या होगा?
If the hypotenuse at a step is \(\sqrt{5}\) and a (1) unit perpendicular is added, what will be the next hypotenuse?
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A \(\sqrt{6}\)
B \(\sqrt{7}\)
C \(\sqrt{10}\)
D \(\sqrt{4}\)
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Correct Answer
A. \(\sqrt{6}\)
Step 1
Concept
The new hypotenuse is \(\sqrt{5+1}=\sqrt{6}\). In a square root spiral, the number increases by one.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{6}\). The new hypotenuse is \(\sqrt{5+1}=\sqrt{6}\). In a square root spiral, the number increases by one.
Step 3
Exam Tip
नया कर्ण \(\sqrt{5+1}=\sqrt{6}\) होगा। वर्गमूल सर्पिल में संख्या एक-एक बढ़ती है।
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वर्गमूल सर्पिल में \(\sqrt{1}\) से शुरू करने पर तीसरा कर्ण कौन-सा होता है?
Starting from \(\sqrt{1}\) in a square root spiral, which is the third hypotenuse?
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A \(\sqrt{2}\)
B \(\sqrt{3}\)
C \(\sqrt{4}\)
D \(\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{3}\)
Step 1
Concept
The sequence of hypotenuses is \(\sqrt{1},\sqrt{2},\sqrt{3}\). Therefore the third hypotenuse is \(\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{3}\). The sequence of hypotenuses is \(\sqrt{1},\sqrt{2},\sqrt{3}\). Therefore the third hypotenuse is \(\sqrt{3}\).
Step 3
Exam Tip
कर्णों का क्रम \(\sqrt{1},\sqrt{2},\sqrt{3}\) होता है। इसलिए तीसरा कर्ण \(\sqrt{3}\) है।
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वर्गमूल सर्पिल में \(\sqrt{4}\) की लंबाई वास्तव में किस पूर्ण संख्या के बराबर है?
In a square root spiral, the length \(\sqrt{4}\) is actually equal to which whole number?
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A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{4}=2\). Not all square roots are irrational.
Step 2
Why this answer is correct
The correct answer is B. (2). \(\sqrt{4}=2\). Not all square roots are irrational.
Step 3
Exam Tip
\(\sqrt{4}=2\) होता है। सभी वर्गमूल अपरिमेय नहीं होते।
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वर्गमूल सर्पिल संख्या रेखा पर किस प्रकार की संख्याओं को दर्शाने में सहायक है?
A square root spiral helps to represent what kind of numbers on the number line?
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A केवल ऋणात्मक संख्याएँ / Only negative numbers
B वर्गमूल वाली संख्याएँ / Numbers involving square roots
C केवल शून्य / Only zero
D केवल सम संख्याएँ / Only even numbers
Explanation opens after your attempt
Correct Answer
B. वर्गमूल वाली संख्याएँ / Numbers involving square roots
Step 1
Concept
A square root spiral helps construct lengths like \(\sqrt{2},\sqrt{3},\sqrt{5}\). These can be shown on the number line.
Step 2
Why this answer is correct
The correct answer is B. वर्गमूल वाली संख्याएँ / Numbers involving square roots. A square root spiral helps construct lengths like \(\sqrt{2},\sqrt{3},\sqrt{5}\). These can be shown on the number line.
Step 3
Exam Tip
वर्गमूल सर्पिल \(\sqrt{2},\sqrt{3},\sqrt{5}\) जैसी लंबाइयाँ बनाने में सहायक है। इससे संख्या रेखा पर स्थान दिखाया जा सकता है।
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वर्गमूल सर्पिल में \(\sqrt{9}\) कर्ण की लंबाई किसके बराबर होगी?
In a square root spiral, the hypotenuse length \(\sqrt{9}\) will be equal to what?
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A (2)
B (3)
C (4)
D (5)
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Step 1
Concept
\(\sqrt{9}=3\). Square roots of perfect squares give whole numbers.
Step 2
Why this answer is correct
The correct answer is B. (3). \(\sqrt{9}=3\). Square roots of perfect squares give whole numbers.
Step 3
Exam Tip
\(\sqrt{9}=3\) होता है। पूर्ण वर्गों के वर्गमूल पूर्ण संख्याएँ देते हैं।
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वर्गमूल सर्पिल में \(\sqrt{8}\) के बाद (1) इकाई लंब जोड़ने पर अगला कर्ण कौन-सा होगा?
In a square root spiral, after \(\sqrt{8}\), which hypotenuse is obtained by adding a (1) unit perpendicular?
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A \(\sqrt{7}\)
B \(\sqrt{8}\)
C \(\sqrt{9}\)
D \(\sqrt{10}\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{9}\)
Step 1
Concept
The new hypotenuse is \(\sqrt{8+1}=\sqrt{9}\). The next step always gives the square root of the next integer.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{9}\). The new hypotenuse is \(\sqrt{8+1}=\sqrt{9}\). The next step always gives the square root of the next integer.
Step 3
Exam Tip
नया कर्ण \(\sqrt{8+1}=\sqrt{9}\) होगा। अगला चरण हमेशा अगले पूर्णांक का वर्गमूल देता है।
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वर्गमूल सर्पिल बनाते समय (1) इकाई की लंब रेखा किस कोण पर खींची जाती है?
While making a square root spiral, at what angle is the (1) unit perpendicular line drawn?
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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
The perpendicular line is drawn at \(90^\circ\). This forms a right triangle.
Step 2
Why this answer is correct
The correct answer is D. \(90^\circ\). The perpendicular line is drawn at \(90^\circ\). This forms a right triangle.
Step 3
Exam Tip
लंब रेखा \(90^\circ\) पर खींची जाती है। इसी से समकोण त्रिभुज बनता है।
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वर्गमूल सर्पिल में \(\sqrt{10}\) लंबाई बनाने के लिए किस पिछले कर्ण पर (1) इकाई लंब जोड़ी जाती है?
To construct length \(\sqrt{10}\) in a square root spiral, a (1) unit perpendicular is added to which previous hypotenuse?
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A \(\sqrt{8}\)
B \(\sqrt{9}\)
C \(\sqrt{10}\)
D \(\sqrt{11}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{9}\)
Step 1
Concept
Adding (1) unit to \(\sqrt{9}\) gives the new hypotenuse \(\sqrt{10}\). Identify the previous square root.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{9}\). Adding (1) unit to \(\sqrt{9}\) gives the new hypotenuse \(\sqrt{10}\). Identify the previous square root.
Step 3
Exam Tip
\(\sqrt{9}\) पर (1) इकाई जोड़ने से नया कर्ण \(\sqrt{10}\) बनता है। पिछले वर्गमूल को पहचानें।
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वर्गमूल सर्पिल में हर नया कर्ण पिछले कर्ण और किस भुजा से बनता है?
In a square root spiral, each new hypotenuse is formed from the previous hypotenuse and which side?
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A (1) इकाई लंब भुजा / (1) unit perpendicular side
B (2) इकाई आधार / (2) unit base
C \(\sqrt{2}\) इकाई व्यास / \(\sqrt{2}\) unit diameter
D (0) इकाई रेखा / (0) unit line
Explanation opens after your attempt
Correct Answer
A. (1) इकाई लंब भुजा / (1) unit perpendicular side
Step 1
Concept
At each step, the previous hypotenuse becomes one side and (1) unit perpendicular is the other side. The new hypotenuse gives the next square root.
Step 2
Why this answer is correct
The correct answer is A. (1) इकाई लंब भुजा / (1) unit perpendicular side. At each step, the previous hypotenuse becomes one side and (1) unit perpendicular is the other side. The new hypotenuse gives the next square root.
Step 3
Exam Tip
हर चरण में पिछला कर्ण एक भुजा बनता है और (1) इकाई लंब दूसरी भुजा। नया कर्ण अगला वर्गमूल देता है।
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\(\sqrt{2}\) को संख्या रेखा पर दिखाने के लिए वर्गमूल सर्पिल से कौन-सी लंबाई कंपास में ली जाती है?
To show \(\sqrt{2}\) on the number line using a square root spiral, which length is taken in the compass?
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A \(\sqrt{2}\) का कर्ण / Hypotenuse of \(\sqrt{2}\)
B \(\sqrt{3}\) का कर्ण / Hypotenuse of \(\sqrt{3}\)
C (2) इकाई लंब / (2) unit perpendicular
D (0) लंबाई / (0) length
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{2}\) का कर्ण / Hypotenuse of \(\sqrt{2}\)
Step 1
Concept
The hypotenuse of length \(\sqrt{2}\) is taken in the compass and an arc is drawn on the number line. This locates the point.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}\) का कर्ण / Hypotenuse of \(\sqrt{2}\). The hypotenuse of length \(\sqrt{2}\) is taken in the compass and an arc is drawn on the number line. This locates the point.
Step 3
Exam Tip
\(\sqrt{2}\) वाला कर्ण कंपास में लेकर संख्या रेखा पर चाप लगाया जाता है। इससे स्थान मिलता है।
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वर्गमूल सर्पिल का दूसरा कर्ण \(\sqrt{2}\) है। चौथा कर्ण कौन-सा होगा?
The second hypotenuse of a square root spiral is \(\sqrt{2}\). What will be the fourth hypotenuse?
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A \(\sqrt{3}\)
B \(\sqrt{4}\)
C \(\sqrt{5}\)
D \(\sqrt{6}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{4}\)
Step 1
Concept
The sequence is \(\sqrt{1},\sqrt{2},\sqrt{3},\sqrt{4}\). Therefore the fourth hypotenuse is \(\sqrt{4}\).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{4}\). The sequence is \(\sqrt{1},\sqrt{2},\sqrt{3},\sqrt{4}\). Therefore the fourth hypotenuse is \(\sqrt{4}\).
Step 3
Exam Tip
कर्णों का क्रम \(\sqrt{1},\sqrt{2},\sqrt{3},\sqrt{4}\) है। इसलिए चौथा कर्ण \(\sqrt{4}\) है।
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वर्गमूल सर्पिल में \(\sqrt{6}\) की लंबाई बनाने से ठीक पहले कौन-सा कर्ण बना होता है?
Just before constructing length \(\sqrt{6}\) in a square root spiral, which hypotenuse has been constructed?
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A \(\sqrt{4}\)
B \(\sqrt{5}\)
C \(\sqrt{6}\)
D \(\sqrt{7}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{5}\)
Step 1
Concept
Adding a (1) unit perpendicular to \(\sqrt{5}\) gives \(\sqrt{6}\). Look at the previous step.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{5}\). Adding a (1) unit perpendicular to \(\sqrt{5}\) gives \(\sqrt{6}\). Look at the previous step.
Step 3
Exam Tip
\(\sqrt{5}\) पर (1) इकाई लंब जोड़ने से \(\sqrt{6}\) मिलता है। पिछले चरण को देखें।
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वर्गमूल सर्पिल में \(\sqrt{7}\) कर्ण के साथ (1) इकाई लंब जोड़ने पर नया कर्ण क्या होगा?
In a square root spiral, if a (1) unit perpendicular is added to the hypotenuse \(\sqrt{7}\), what will be the new hypotenuse?
#number-systems
#square-root-spiral
#next-root
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A \(\sqrt{6}\)
B \(\sqrt{7}\)
C \(\sqrt{8}\)
D \(\sqrt{9}\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{8}\)
Step 1
Concept
The new hypotenuse is \(\sqrt{7+1}=\sqrt{8}\). Each new step adds (1) to the number.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{8}\). The new hypotenuse is \(\sqrt{7+1}=\sqrt{8}\). Each new step adds (1) to the number.
Step 3
Exam Tip
नया कर्ण \(\sqrt{7+1}=\sqrt{8}\) होगा। हर नया चरण संख्या में (1) जोड़ता है।
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वर्गमूल सर्पिल में \(\sqrt{5}\) कौन-से प्रकार की संख्या का उदाहरण है?
In a square root spiral, \(\sqrt{5}\) is an example of which type of number?
#number-systems
#square-root-spiral
#irrational-number
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A पूर्ण संख्या / Whole number
B अपरिमेय संख्या / Irrational number
C शून्य / Zero
D ऋणात्मक पूर्णांक / Negative integer
Explanation opens after your attempt
Correct Answer
B. 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 B. 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 \(\sqrt{16}\) की लंबाई किस संख्या के बराबर होगी?
In a square root spiral, the length \(\sqrt{16}\) will be equal to which number?
#number-systems
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#perfect-square
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A (2)
B (3)
C (4)
D (5)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{16}=4\). Square roots of perfect squares are rational and whole numbers.
Step 2
Why this answer is correct
The correct answer is C. (4). \(\sqrt{16}=4\). Square roots of perfect squares are rational and whole numbers.
Step 3
Exam Tip
\(\sqrt{16}=4\) होता है। पूर्ण वर्गों के वर्गमूल परिमेय और पूर्ण संख्या होते हैं।
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वर्गमूल सर्पिल में \(\sqrt{11}\) बनाने के लिए पिछले कर्ण की लंबाई क्या होनी चाहिए?
To make \(\sqrt{11}\) in a square root spiral, what should be the previous hypotenuse length?
#number-systems
#square-root-spiral
#previous-root
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A \(\sqrt{9}\)
B \(\sqrt{10}\)
C \(\sqrt{11}\)
D \(\sqrt{12}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{10}\)
Step 1
Concept
Adding a (1) unit perpendicular to \(\sqrt{10}\) forms \(\sqrt{11}\). Remember the stepwise construction.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{10}\). Adding a (1) unit perpendicular to \(\sqrt{10}\) forms \(\sqrt{11}\). Remember the stepwise construction.
Step 3
Exam Tip
\(\sqrt{10}\) के साथ (1) इकाई लंब जोड़ने पर \(\sqrt{11}\) बनता है। क्रमिक निर्माण याद रखें।
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वर्गमूल सर्पिल बनाते समय कर्ण को संख्या रेखा पर स्थानांतरित करने के लिए कौन-सा उपकरण उपयोगी है?
While making a square root spiral, which tool is useful to transfer the hypotenuse length to the number line?
#number-systems
#square-root-spiral
#compass
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A कंपास / Compass
B तराजू / Balance
C घड़ी / Clock
D कैलकुलेटर / Calculator
Explanation opens after your attempt
Correct Answer
A. कंपास / Compass
Step 1
Concept
With a compass, the hypotenuse length is taken and an arc is drawn on the number line. This is the geometric construction.
Step 2
Why this answer is correct
The correct answer is A. कंपास / Compass. With a compass, the hypotenuse length is taken and an arc is drawn on the number line. This is the geometric construction.
Step 3
Exam Tip
कंपास से कर्ण की लंबाई लेकर संख्या रेखा पर चाप लगाया जाता है। यही ज्यामितीय निर्माण है।
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वर्गमूल सर्पिल में \(\sqrt{2}\) से \(\sqrt{3}\) बनने का कारण कौन-सा है?
What is the reason for \(\sqrt{2}\) becoming \(\sqrt{3}\) in a square root spiral?
#number-systems
#square-root-spiral
#pythagoras
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A (\(\sqrt{2}\)2 +12 =3)
B (\(\sqrt{2}\)2 +22 =6)
C (\(\sqrt{2}\)2 -12 =1)
D \(\sqrt{2}+1=\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
A. (\(\sqrt{2}\)2 +12 =3)
Step 1
Concept
By Pythagoras theorem, the new hypotenuse becomes \(\sqrt{2+1}=\sqrt{3}\). Do not treat \(\sqrt{2}+1\) as \(\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. (\(\sqrt{2}\)2 +12 =3). By Pythagoras theorem, the new hypotenuse becomes \(\sqrt{2+1}=\sqrt{3}\). Do not treat \(\sqrt{2}+1\) as \(\sqrt{3}\).
Step 3
Exam Tip
पाइथागोरस प्रमेय से नया कर्ण \(\sqrt{2+1}=\sqrt{3}\) बनता है। \(\sqrt{2}+1\) को \(\sqrt{3}\) न मानें।
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वर्गमूल सर्पिल में \(\sqrt{3}\) से \(\sqrt{4}\) बनने में किस गणना का प्रयोग होता है?
Which calculation is used when \(\sqrt{3}\) becomes \(\sqrt{4}\) in a square root spiral?
#number-systems
#square-root-spiral
#pythagoras
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A (\(\sqrt{3}\)2 +12 =4)
B (\(\sqrt{3}\)2 +32 =12)
C \(\sqrt{3}+1=\sqrt{4}\)
D (\(\sqrt{3}\)2 -12 =2)
Explanation opens after your attempt
Correct Answer
A. (\(\sqrt{3}\)2 +12 =4)
Step 1
Concept
With \(\sqrt{3}\) and (1) unit, the new hypotenuse is \(\sqrt{4}\). The correct calculation uses sum of squares.
Step 2
Why this answer is correct
The correct answer is A. (\(\sqrt{3}\)2 +12 =4). With \(\sqrt{3}\) and (1) unit, the new hypotenuse is \(\sqrt{4}\). The correct calculation uses sum of squares.
Step 3
Exam Tip
\(\sqrt{3}\) और (1) इकाई से नया कर्ण \(\sqrt{4}\) मिलता है। सही गणना वर्गों के योग से होती है।
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वर्गमूल सर्पिल में बनने वाली रेखाएँ धीरे-धीरे किस आकृति जैसी दिखती हैं?
The lines formed in a square root spiral gradually look like which shape?
#number-systems
#square-root-spiral
#shape
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A सर्पिल / Spiral
B वर्ग / Square
C आयत / Rectangle
D सीधी रेखा / Straight line
Explanation opens after your attempt
Correct Answer
A. सर्पिल / Spiral
Step 1
Concept
Successive right triangles turn around and make a spiral-like shape. Hence it is called a square root spiral.
Step 2
Why this answer is correct
The correct answer is A. सर्पिल / Spiral. Successive right triangles turn around and make a spiral-like shape. Hence it is called a square root spiral.
Step 3
Exam Tip
लगातार बने समकोण त्रिभुज मुड़ते हुए सर्पिल जैसी आकृति बनाते हैं। इसलिए इसे वर्गमूल सर्पिल कहते हैं।
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वर्गमूल सर्पिल में \(\sqrt{1}\) की लंबाई किसके बराबर है?
In a square root spiral, the length \(\sqrt{1}\) is equal to what?
#number-systems
#square-root-spiral
#start
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A (0)
B (1)
C (2)
D \(\sqrt{2}\)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{1}=1\). This is the initial (1) unit line segment.
Step 2
Why this answer is correct
The correct answer is B. (1). \(\sqrt{1}=1\). This is the initial (1) unit line segment.
Step 3
Exam Tip
\(\sqrt{1}=1\) होता है। यही प्रारंभिक (1) इकाई रेखाखंड है।
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वर्गमूल सर्पिल में \(\sqrt{13}\) बनाने से ठीक पहले कौन-सा कर्ण होना चाहिए?
Just before making \(\sqrt{13}\) in a square root spiral, which hypotenuse should be present?
#number-systems
#square-root-spiral
#previous-root
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A \(\sqrt{11}\)
B \(\sqrt{12}\)
C \(\sqrt{13}\)
D \(\sqrt{14}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{12}\)
Step 1
Concept
Adding a (1) unit perpendicular to \(\sqrt{12}\) forms \(\sqrt{13}\). The previous hypotenuse has one less number.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{12}\). Adding a (1) unit perpendicular to \(\sqrt{12}\) forms \(\sqrt{13}\). The previous hypotenuse has one less number.
Step 3
Exam Tip
\(\sqrt{12}\) में (1) इकाई लंब जोड़ने पर नया कर्ण \(\sqrt{13}\) बनता है। पिछला कर्ण एक कम संख्या का होता है।
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वर्गमूल सर्पिल में \(\sqrt{14}\) कर्ण के बाद अगला कर्ण कौन-सा होगा?
In a square root spiral, what will be the next hypotenuse after \(\sqrt{14}\)?
#number-systems
#square-root-spiral
#next-root
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A \(\sqrt{13}\)
B \(\sqrt{14}\)
C \(\sqrt{15}\)
D \(\sqrt{16}\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{15}\)
Step 1
Concept
At each new step, the number increases by (1). Therefore \(\sqrt{15}\) comes after \(\sqrt{14}\).
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{15}\). At each new step, the number increases by (1). Therefore \(\sqrt{15}\) comes after \(\sqrt{14}\).
Step 3
Exam Tip
हर नए चरण में संख्या (1) बढ़ती है। इसलिए \(\sqrt{14}\) के बाद \(\sqrt{15}\) आता है।
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वर्गमूल सर्पिल में किसी चरण पर पिछला कर्ण \(\sqrt{n}\) हो, तो (1) इकाई लंब जोड़ने पर नया कर्ण क्या होगा?
If the previous hypotenuse at a step in a square root spiral is \(\sqrt{n}\), what is the new hypotenuse after adding a (1) unit perpendicular?
#number-systems
#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
A. \(\sqrt{n+1}\)
Step 1
Concept
The new hypotenuse is (\sqrt{\(\sqrt{n}\)2 +12 }=\sqrt{n+1}). This is the rule of the square root spiral.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{n+1}\). The new hypotenuse is (\sqrt{\(\sqrt{n}\)2 +12 }=\sqrt{n+1}). This is the rule of the square root spiral.
Step 3
Exam Tip
नया कर्ण (\sqrt{\(\sqrt{n}\)2 +12 }=\sqrt{n+1}) होता है। यह वर्गमूल सर्पिल का नियम है।
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वर्गमूल सर्पिल में (1) इकाई लंब जोड़ने का उद्देश्य क्या है?
What is the purpose of adding a (1) unit perpendicular in a square root spiral?
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#purpose
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A अगला वर्गमूल कर्ण बनाना / To make the next square root as hypotenuse
B रेखा को मिटाना / To erase the line
C हर को शून्य करना / To make denominator zero
D त्रिभुज को वृत्त बनाना / To make triangle a circle
Explanation opens after your attempt
Correct Answer
A. अगला वर्गमूल कर्ण बनाना / To make the next square root as hypotenuse
Step 1
Concept
Adding a (1) unit perpendicular forms a new right triangle. Its hypotenuse gives the next square root.
Step 2
Why this answer is correct
The correct answer is A. अगला वर्गमूल कर्ण बनाना / To make the next square root as hypotenuse. Adding a (1) unit perpendicular forms a new right triangle. Its hypotenuse gives the next square root.
Step 3
Exam Tip
(1) इकाई लंब जोड़ने से नया समकोण त्रिभुज बनता है। उसका कर्ण अगला वर्गमूल देता है।
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वर्गमूल सर्पिल में \(\sqrt{2}\) की लंबाई किस दो भुजाओं वाले समकोण त्रिभुज से मिलती है?
In a square root spiral, length \(\sqrt{2}\) is obtained from a right triangle with which two sides?
#number-systems
#square-root-spiral
#sqrt2
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A (1) इकाई और (1) इकाई / (1) unit and (1) unit
B (2) इकाई और (2) इकाई / (2) units and (2) units
C (1) इकाई और (2) इकाई / (1) unit and (2) units
D (3) इकाई और (1) इकाई / (3) units and (1) unit
Explanation opens after your attempt
Correct Answer
A. (1) इकाई और (1) इकाई / (1) unit and (1) unit
Step 1
Concept
Sides (1) and (1) units give hypotenuse \(\sqrt{2}\). This is the first main step of the spiral.
Step 2
Why this answer is correct
The correct answer is A. (1) इकाई और (1) इकाई / (1) unit and (1) unit. Sides (1) and (1) units give hypotenuse \(\sqrt{2}\). This is the first main step of the spiral.
Step 3
Exam Tip
(1) और (1) इकाई भुजाओं से कर्ण \(\sqrt{2}\) मिलता है। यह सर्पिल का पहला मुख्य चरण है।
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वर्गमूल सर्पिल में \(\sqrt{3}\) बनाने के लिए कौन-सी दो भुजाएँ समकोण बनाती हैं?
In a square root spiral, which two sides form the right angle to make \(\sqrt{3}\)?
#number-systems
#square-root-spiral
#sqrt3
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A \(\sqrt{2}\) और (1) / \(\sqrt{2}\) and (1)
B \(\sqrt{3}\) और (1) / \(\sqrt{3}\) and (1)
C (2) और (2) / (2) and (2)
D (3) और (1) / (3) and (1)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{2}\) और (1) / \(\sqrt{2}\) and (1)
Step 1
Concept
\(\sqrt{2}\) is the previous hypotenuse and (1) unit is the new perpendicular. Together they form new hypotenuse \(\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}\) और (1) / \(\sqrt{2}\) and (1). \(\sqrt{2}\) is the previous hypotenuse and (1) unit is the new perpendicular. Together they form new hypotenuse \(\sqrt{3}\).
Step 3
Exam Tip
\(\sqrt{2}\) पिछला कर्ण है और (1) इकाई नई लंब है। इनसे नया कर्ण \(\sqrt{3}\) बनता है।
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वर्गमूल सर्पिल में \(\sqrt{4}\) बनाने के बाद अगली अपरिमेय लंबाई कौन-सी हो सकती है?
After making \(\sqrt{4}\) in a square root spiral, which next irrational length can be obtained?
#number-systems
#square-root-spiral
#irrational-length
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A \(\sqrt{5}\)
B \(\sqrt{4}\)
C (2)
D \(\sqrt{9}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{5}\)
Step 1
Concept
\(\sqrt{4}=2\) is rational. The next hypotenuse is \(\sqrt{5}\), which is irrational.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{5}\). \(\sqrt{4}=2\) is rational. The next hypotenuse is \(\sqrt{5}\), which is irrational.
Step 3
Exam Tip
\(\sqrt{4}=2\) परिमेय है। अगला कर्ण \(\sqrt{5}\) होगा, जो अपरिमेय है।
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वर्गमूल सर्पिल में \(\sqrt{25}\) जैसी लंबाई किस प्रकार की संख्या देती है?
In a square root spiral, a length like \(\sqrt{25}\) gives which type of number?
#number-systems
#square-root-spiral
#perfect-square
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A पूर्ण संख्या / Whole number
B अपरिमेय संख्या / Irrational number
C ऋणात्मक संख्या / Negative number
D अपरिभाषित संख्या / Undefined number
Explanation opens after your attempt
Correct Answer
A. पूर्ण संख्या / Whole number
Step 1
Concept
\(\sqrt{25}=5\). Square roots of perfect squares give whole numbers.
Step 2
Why this answer is correct
The correct answer is A. पूर्ण संख्या / Whole number. \(\sqrt{25}=5\). Square roots of perfect squares give whole numbers.
Step 3
Exam Tip
\(\sqrt{25}=5\) होता है। पूर्ण वर्गों के वर्गमूल पूर्ण संख्या देते हैं।
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वर्गमूल सर्पिल में \(\sqrt{12}\) के बाद अगला कर्ण कौन-सा बनेगा?
In a square root spiral, which hypotenuse will be formed after \(\sqrt{12}\)?
#number-systems
#square-root-spiral
#next-root
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A \(\sqrt{11}\)
B \(\sqrt{12}\)
C \(\sqrt{13}\)
D \(\sqrt{14}\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{13}\)
Step 1
Concept
The new hypotenuse is \(\sqrt{12+1}=\sqrt{13}\). Take the next square root in sequence.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{13}\). The new hypotenuse is \(\sqrt{12+1}=\sqrt{13}\). Take the next square root in sequence.
Step 3
Exam Tip
नया कर्ण \(\sqrt{12+1}=\sqrt{13}\) होता है। क्रम में अगला वर्गमूल लें।
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वर्गमूल सर्पिल में \(\sqrt{18}\) बनाने के लिए किस पिछले कर्ण पर (1) इकाई लंब बनाई जाएगी?
To make \(\sqrt{18}\) in a square root spiral, on which previous hypotenuse is a (1) unit perpendicular drawn?
#number-systems
#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
With \(\sqrt{17}\) and a (1) unit perpendicular, \(\sqrt{18}\) is formed. The previous number is always (1) less.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{17}\). With \(\sqrt{17}\) and a (1) unit perpendicular, \(\sqrt{18}\) is formed. The previous number is always (1) less.
Step 3
Exam Tip
\(\sqrt{17}\) के साथ (1) इकाई लंब से \(\sqrt{18}\) बनता है। पिछली संख्या हमेशा (1) कम होती है।
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वर्गमूल सर्पिल में \(\sqrt{20}\) की लंबाई संख्या रेखा पर रखने के लिए किस लंबाई को कंपास में लेना होगा?
To place length \(\sqrt{20}\) on the number line using a square root spiral, which length must be taken in the compass?
#number-systems
#square-root-spiral
#number-line
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A \(\sqrt{19}\) का कर्ण / Hypotenuse of \(\sqrt{19}\)
B \(\sqrt{20}\) का कर्ण / Hypotenuse of \(\sqrt{20}\)
C (20) इकाई / (20) units
D (1) इकाई / (1) unit
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{20}\) का कर्ण / Hypotenuse of \(\sqrt{20}\)
Step 1
Concept
The hypotenuse length of the required square root is taken in the compass to place it on the number line.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{20}\) का कर्ण / Hypotenuse of \(\sqrt{20}\). The hypotenuse length of the required square root is taken in the compass to place it on the number line.
Step 3
Exam Tip
जिस वर्गमूल को संख्या रेखा पर रखना है, उसी कर्ण की लंबाई कंपास में ली जाती है।
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वर्गमूल सर्पिल में कौन-सा कथन गलत है?
Which statement is incorrect for a square root spiral?
#number-systems
#square-root-spiral
#concept-check
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A हर नया त्रिभुज समकोण त्रिभुज होता है / Each new triangle is a right triangle
B हर नई लंब भुजा (1) इकाई होती है / Each new perpendicular side is (1) unit
C कर्ण क्रम में \(\sqrt{1},\sqrt{2},\sqrt{3}\) आते हैं / Hypotenuses come as \(\sqrt{1},\sqrt{2},\sqrt{3}\)
D हर कर्ण हमेशा पूर्ण संख्या होता है / Every hypotenuse is always a whole number
Explanation opens after your attempt
Correct Answer
D. हर कर्ण हमेशा पूर्ण संख्या होता है / Every hypotenuse is always a whole number
Step 1
Concept
Many hypotenuses like \(\sqrt{2},\sqrt{3},\sqrt{5}\) are not whole numbers. So this statement is incorrect.
Step 2
Why this answer is correct
The correct answer is D. हर कर्ण हमेशा पूर्ण संख्या होता है / Every hypotenuse is always a whole number. Many hypotenuses like \(\sqrt{2},\sqrt{3},\sqrt{5}\) are not whole numbers. So this statement is incorrect.
Step 3
Exam Tip
\(\sqrt{2},\sqrt{3},\sqrt{5}\) जैसे कई कर्ण पूर्ण संख्या नहीं होते। इसलिए यह कथन गलत है।
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वर्गमूल सर्पिल में \(\sqrt{2}\) और \(\sqrt{3}\) के बीच कौन-सा संबंध सही है?
Which relation between \(\sqrt{2}\) and \(\sqrt{3}\) is correct in a square root spiral?
#number-systems
#square-root-spiral
#order
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A \(\sqrt{3}\) अगला कर्ण है / \(\sqrt{3}\) is the next hypotenuse
B \(\sqrt{2}\) और \(\sqrt{3}\) बराबर हैं / \(\sqrt{2}\) and \(\sqrt{3}\) are equal
C \(\sqrt{3}\) पहले आता है / \(\sqrt{3}\) comes first
D \(\sqrt{2}\) नहीं बनता / \(\sqrt{2}\) is not formed
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{3}\) अगला कर्ण है / \(\sqrt{3}\) is the next hypotenuse
Step 1
Concept
After \(\sqrt{2}\), adding a (1) unit perpendicular forms \(\sqrt{3}\). Keep the order in mind.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{3}\) अगला कर्ण है / \(\sqrt{3}\) is the next hypotenuse. After \(\sqrt{2}\), adding a (1) unit perpendicular forms \(\sqrt{3}\). Keep the order in mind.
Step 3
Exam Tip
\(\sqrt{2}\) के बाद (1) इकाई लंब जोड़ने से \(\sqrt{3}\) बनता है। क्रम को ध्यान रखें।
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वर्गमूल सर्पिल में \(\sqrt{24}\) के बाद कौन-सा कर्ण आएगा?
In a square root spiral, which hypotenuse comes after \(\sqrt{24}\)?
#number-systems
#square-root-spiral
#next-root
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A \(\sqrt{23}\)
B \(\sqrt{24}\)
C \(\sqrt{25}\)
D \(\sqrt{26}\)
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Correct Answer
C. \(\sqrt{25}\)
Step 1
Concept
At each next step, the number increases by (1). Therefore \(\sqrt{25}\) comes after \(\sqrt{24}\).
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{25}\). At each next step, the number increases by (1). Therefore \(\sqrt{25}\) comes after \(\sqrt{24}\).
Step 3
Exam Tip
हर अगले चरण में संख्या (1) बढ़ती है। इसलिए \(\sqrt{24}\) के बाद \(\sqrt{25}\) आता है।
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वर्गमूल सर्पिल में \(\sqrt{15}\) बनाने से ठीक पहले कौन-सी लंबाई बन चुकी होगी?
Just before making \(\sqrt{15}\) in a square root spiral, which length will have already been made?
#number-systems
#square-root-spiral
#previous-root
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A \(\sqrt{13}\)
B \(\sqrt{14}\)
C \(\sqrt{15}\)
D \(\sqrt{16}\)
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Correct Answer
B. \(\sqrt{14}\)
Step 1
Concept
\(\sqrt{14}\) will be the previous hypotenuse. Adding a (1) unit perpendicular to it gives \(\sqrt{15}\).
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{14}\). \(\sqrt{14}\) will be the previous hypotenuse. Adding a (1) unit perpendicular to it gives \(\sqrt{15}\).
Step 3
Exam Tip
\(\sqrt{14}\) पिछले चरण का कर्ण होगा। उसमें (1) इकाई लंब जोड़कर \(\sqrt{15}\) बनेगा।
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वर्गमूल सर्पिल का मुख्य ज्यामितीय लाभ क्या है?
What is the main geometric benefit of a square root spiral?
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#square-root-spiral
#benefit
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A वर्गमूल लंबाइयों का निर्माण करना / To construct square root lengths
B सिर्फ वृत्त का क्षेत्रफल निकालना / To find only area of circle
C सिर्फ त्रिभुज का रंग भरना / To only color triangles
D हर को शून्य करना / To make denominator zero
Explanation opens after your attempt
Correct Answer
A. वर्गमूल लंबाइयों का निर्माण करना / To construct square root lengths
Step 1
Concept
A square root spiral geometrically constructs lengths like \(\sqrt{2},\sqrt{3},\sqrt{4}\). It is also useful on the number line.
Step 2
Why this answer is correct
The correct answer is A. वर्गमूल लंबाइयों का निर्माण करना / To construct square root lengths. A square root spiral geometrically constructs lengths like \(\sqrt{2},\sqrt{3},\sqrt{4}\). It is also useful on the number line.
Step 3
Exam Tip
वर्गमूल सर्पिल \(\sqrt{2},\sqrt{3},\sqrt{4}\) जैसी लंबाइयाँ ज्यामितीय रूप से बनाता है। यह संख्या रेखा पर भी उपयोगी है।
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वर्गमूल सर्पिल में \(\sqrt{2}\) बनाने के लिए किस प्रमेय से \(\sqrt{1^2+1^2}\) लिखा जाता है?
In a square root spiral, which theorem allows writing \(\sqrt{1^2+1^2}\) to make \(\sqrt{2}\)?
#number-systems
#square-root-spiral
#pythagoras
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A पाइथागोरस प्रमेय / Pythagoras theorem
B गुणनखंड प्रमेय / Factor theorem
C शेषफल प्रमेय / Remainder theorem
D कोण द्विभाजक प्रमेय / Angle bisector theorem
Explanation opens after your attempt
Correct Answer
A. पाइथागोरस प्रमेय / Pythagoras theorem
Step 1
Concept
Pythagoras theorem is used for the hypotenuse in a right triangle. Thus \(\sqrt{1^2+1^2}=\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. पाइथागोरस प्रमेय / Pythagoras theorem. Pythagoras theorem is used for the hypotenuse in a right triangle. Thus \(\sqrt{1^2+1^2}=\sqrt{2}\).
Step 3
Exam Tip
समकोण त्रिभुज में कर्ण के लिए पाइथागोरस प्रमेय लगता है। इसलिए \(\sqrt{1^2+1^2}=\sqrt{2}\) मिलता है।
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वर्गमूल सर्पिल में यदि नया कर्ण \(\sqrt{n+1}\) है, तो पिछला कर्ण क्या था?
In a square root spiral, if the new hypotenuse is \(\sqrt{n+1}\), what was the previous hypotenuse?
#number-systems
#square-root-spiral
#general-rule
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A \(\sqrt{n}\)
B \(\sqrt{n+2}\)
C \(\sqrt{2n}\)
D \(\sqrt{n^2}\)
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Correct Answer
A. \(\sqrt{n}\)
Step 1
Concept
The new hypotenuse is formed from previous \(\sqrt{n}\) and a (1) unit perpendicular. Therefore the previous hypotenuse was \(\sqrt{n}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{n}\). The new hypotenuse is formed from previous \(\sqrt{n}\) and a (1) unit perpendicular. Therefore the previous hypotenuse was \(\sqrt{n}\).
Step 3
Exam Tip
नया कर्ण पिछले \(\sqrt{n}\) और (1) इकाई लंब से बनता है। इसलिए पिछला कर्ण \(\sqrt{n}\) था।
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वर्गमूल सर्पिल में \(\sqrt{6}\) और (1) इकाई लंब से नया कर्ण क्या होगा?
In a square root spiral, what will be the new hypotenuse from \(\sqrt{6}\) and a (1) unit perpendicular?
#number-systems
#square-root-spiral
#next-root
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A \(\sqrt{5}\)
B \(\sqrt{6}\)
C \(\sqrt{7}\)
D \(\sqrt{8}\)
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Correct Answer
C. \(\sqrt{7}\)
Step 1
Concept
The new hypotenuse is \(\sqrt{6+1}=\sqrt{7}\). This follows from Pythagoras theorem.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{7}\). The new hypotenuse is \(\sqrt{6+1}=\sqrt{7}\). This follows from Pythagoras theorem.
Step 3
Exam Tip
नया कर्ण \(\sqrt{6+1}=\sqrt{7}\) होगा। यह पाइथागोरस प्रमेय से मिलता है।
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वर्गमूल सर्पिल में \(\sqrt{2}\) की लंबाई लगभग (1) और (2) के बीच क्यों रखी जाती है?
Why is the length \(\sqrt{2}\) placed between (1) and (2) in a square root spiral?
#number-systems
#square-root-spiral
#number-line
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A क्योंकि \(1^2<2<2^2\) / Because \(1^2<2<2^2\)
B क्योंकि (2<1) / Because (2<1)
C क्योंकि \(\sqrt{2}=2\) / Because \(\sqrt{2}=2\)
D क्योंकि \(\sqrt{2}=0\) / Because \(\sqrt{2}=0\)
Explanation opens after your attempt
Correct Answer
A. क्योंकि \(1^2<2<2^2\) / Because \(1^2<2<2^2\)
Step 1
Concept
From \(1^2<2<2^2\), we get \(1<\sqrt{2}<2\). This helps locate it on the number line.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि \(1^2<2<2^2\) / Because \(1^2<2<2^2\). From \(1^2<2<2^2\), we get \(1<\sqrt{2}<2\). This helps locate it on the number line.
Step 3
Exam Tip
\(1^2<2<2^2\) से \(1<\sqrt{2}<2\) मिलता है। संख्या रेखा पर स्थान इसी से समझ आता है।
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वर्गमूल सर्पिल में \(\sqrt{3}\) की लंबाई किन पूर्ण संख्याओं के बीच होगी?
In a square root spiral, the length \(\sqrt{3}\) lies between which whole numbers?
#number-systems
#square-root-spiral
#number-line
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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 \(1<\sqrt{3}<2\).
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 \(1<\sqrt{3}<2\).
Step 3
Exam Tip
क्योंकि \(1^2<3<2^2\) है। इसलिए \(1<\sqrt{3}<2\) होता है।
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वर्गमूल सर्पिल में \(\sqrt{5}\) की लंबाई किन पूर्ण संख्याओं के बीच होगी?
In a square root spiral, the length \(\sqrt{5}\) lies between which whole numbers?
#number-systems
#square-root-spiral
#number-line
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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
B. (2) और (3) / (2) and (3)
Step 1
Concept
Because \(2^2<5<3^2\). Therefore \(2<\sqrt{5}<3\).
Step 2
Why this answer is correct
The correct answer is B. (2) और (3) / (2) and (3). Because \(2^2<5<3^2\). Therefore \(2<\sqrt{5}<3\).
Step 3
Exam Tip
क्योंकि \(2^2<5<3^2\) है। इसलिए \(2<\sqrt{5}<3\) होता है।
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