वर्गमूल सर्पिल का निर्माण किस मूल लंबाई से शुरू किया जाता है?
The construction of a square root spiral starts with which basic length?
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A (1) इकाई / (1) unit
B (2) इकाई / (2) units
C \(\sqrt{2}\) इकाई / \(\sqrt{2}\) units
D \(\sqrt{3}\) इकाई / \(\sqrt{3}\) units
Explanation opens after your attempt
Correct Answer
A. (1) इकाई / (1) unit
Step 1
Concept
A square root spiral starts with a (1) unit line segment. Right triangles are then built from this base.
Step 2
Why this answer is correct
The correct answer is A. (1) इकाई / (1) unit. A square root spiral starts with a (1) unit line segment. Right triangles are then built from this base.
Step 3
Exam Tip
वर्गमूल सर्पिल में शुरुआत (1) इकाई रेखाखंड से होती है। इसी आधार से आगे समकोण त्रिभुज बनते हैं।
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वर्गमूल सर्पिल में पहले समकोण त्रिभुज की दोनों लंब भुजाएँ (1) इकाई हों तो कर्ण कौन-सा होगा?
In a square root spiral, if both perpendicular sides of the first right triangle are (1) unit, what will be the hypotenuse?
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#square-root-spiral
#pythagoras
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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
The hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). This follows from Pythagoras theorem.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{2}\). The hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). This follows from Pythagoras theorem.
Step 3
Exam Tip
कर्ण \(\sqrt{1^2+1^2}=\sqrt{2}\) होता है। यह पाइथागोरस प्रमेय से मिलता है।
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वर्गमूल सर्पिल में प्रत्येक नया समकोण किस प्रकार बनाया जाता है?
How is each new right angle made in a square root spiral?
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A पिछले कर्ण पर (1) इकाई लंब खींचकर / By drawing a (1) unit perpendicular on the previous hypotenuse
B पिछले कर्ण को आधा करके / By halving the previous hypotenuse
C पिछले कर्ण को मिटाकर / By erasing the previous hypotenuse
D हर बार (2) इकाई रेखा जोड़कर / By adding a (2) unit line each time
Explanation opens after your attempt
Correct Answer
A. पिछले कर्ण पर (1) इकाई लंब खींचकर / By drawing a (1) unit perpendicular on the previous hypotenuse
Step 1
Concept
At each new step a (1) unit perpendicular is drawn on the previous hypotenuse. This makes a new right triangle.
Step 2
Why this answer is correct
The correct answer is A. पिछले कर्ण पर (1) इकाई लंब खींचकर / By drawing a (1) unit perpendicular on the previous hypotenuse. At each new step a (1) unit perpendicular is drawn on the previous hypotenuse. This makes a new right triangle.
Step 3
Exam Tip
हर नए चरण में पिछले कर्ण पर (1) इकाई लंब रेखा बनाई जाती है। इससे नया समकोण त्रिभुज बनता है।
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यदि पिछले कर्ण की लंबाई \(\sqrt{4}\) है और (1) इकाई लंब जोड़ी जाए तो नया कर्ण क्या होगा?
If the previous hypotenuse is \(\sqrt{4}\) and a (1) unit perpendicular is added, what will be the new hypotenuse?
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#square-root-spiral
#next-root
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A \(\sqrt{3}\)
B \(\sqrt{4}\)
C \(\sqrt{5}\)
D \(\sqrt{8}\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{5}\)
Step 1
Concept
The new hypotenuse is \(\sqrt{4+1}=\sqrt{5}\). At every step the number increases by (1).
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{5}\). The new hypotenuse is \(\sqrt{4+1}=\sqrt{5}\). At every step the number increases by (1).
Step 3
Exam Tip
नया कर्ण \(\sqrt{4+1}=\sqrt{5}\) होगा। हर चरण में संख्या (1) बढ़ती है।
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वर्गमूल सर्पिल में \(\sqrt{7}\) बनाने से ठीक पहले कौन-सा कर्ण बनता है?
Which hypotenuse is constructed just before \(\sqrt{7}\) in a square root spiral?
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#previous-root
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A \(\sqrt{5}\)
B \(\sqrt{6}\)
C \(\sqrt{7}\)
D \(\sqrt{8}\)
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Correct Answer
B. \(\sqrt{6}\)
Step 1
Concept
Adding a (1) unit perpendicular to \(\sqrt{6}\) gives \(\sqrt{7}\). The previous hypotenuse has one less number.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{6}\). Adding a (1) unit perpendicular to \(\sqrt{6}\) gives \(\sqrt{7}\). The previous hypotenuse has one less number.
Step 3
Exam Tip
\(\sqrt{6}\) में (1) इकाई लंब जोड़ने पर \(\sqrt{7}\) मिलता है। पिछला कर्ण हमेशा एक कम संख्या का होता है।
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वर्गमूल सर्पिल में कर्णों का प्रारंभिक सही क्रम कौन-सा है?
What is the correct initial order of hypotenuses in a square root spiral?
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A \(\sqrt{1},\sqrt{2},\sqrt{3},\sqrt{4}\)
B \(\sqrt{1},\sqrt{3},\sqrt{2},\sqrt{4}\)
C \(\sqrt{2},\sqrt{4},\sqrt{6},\sqrt{8}\)
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. Recognizing the order is useful in easy questions.
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. Recognizing the order is useful in easy questions.
Step 3
Exam Tip
सर्पिल में कर्ण क्रम से \(\sqrt{1},\sqrt{2},\sqrt{3}\) आगे बढ़ते हैं। क्रम पहचानना आसान प्रश्नों में बहुत उपयोगी है।
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वर्गमूल सर्पिल में \(\sqrt{6}\) के बाद कौन-सी लंबाई आती है?
Which length comes after \(\sqrt{6}\) in a square root spiral?
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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
At each new step the next square root appears. Therefore \(\sqrt{7}\) comes after \(\sqrt{6}\).
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{7}\). At each new step the next square root appears. Therefore \(\sqrt{7}\) comes after \(\sqrt{6}\).
Step 3
Exam Tip
हर नए चरण में अगला वर्गमूल आता है। इसलिए \(\sqrt{6}\) के बाद \(\sqrt{7}\) आता है।
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वर्गमूल सर्पिल में कौन-सा प्रमेय बार-बार प्रयोग होता है?
Which theorem is repeatedly used in a square root spiral?
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#pythagoras
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A पाइथागोरस प्रमेय / Pythagoras theorem
B शेषफल प्रमेय / Remainder theorem
C गुणनखंड प्रमेय / Factor theorem
D थेल्स का विपरीत प्रमेय / Converse of Thales theorem
Explanation opens after your attempt
Correct Answer
A. पाइथागोरस प्रमेय / Pythagoras theorem
Step 1
Concept
Each step is based on a right triangle. Therefore Pythagoras theorem is used to find the hypotenuse.
Step 2
Why this answer is correct
The correct answer is A. पाइथागोरस प्रमेय / Pythagoras theorem. Each step is based on a right triangle. Therefore Pythagoras theorem is used to find the hypotenuse.
Step 3
Exam Tip
हर चरण समकोण त्रिभुज पर आधारित है। इसलिए कर्ण निकालने के लिए पाइथागोरस प्रमेय प्रयोग होता है।
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वर्गमूल सर्पिल में \(\sqrt{9}\) की लंबाई किस पूर्ण संख्या के बराबर है?
In a square root spiral, the length \(\sqrt{9}\) equals which whole number?
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A (2)
B (3)
C (4)
D (9)
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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 \(\sqrt{10}\) बनाने के लिए किस पिछले कर्ण पर (1) इकाई लंब बनाई जाती है?
To make \(\sqrt{10}\) in a square root spiral, a (1) unit perpendicular is drawn on 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
The previous hypotenuse \(\sqrt{9}\) with a (1) unit perpendicular gives \(\sqrt{10}\). Identifying the previous step is important.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{9}\). The previous hypotenuse \(\sqrt{9}\) with a (1) unit perpendicular gives \(\sqrt{10}\). Identifying the previous step is important.
Step 3
Exam Tip
\(\sqrt{9}\) और (1) इकाई लंब से नया कर्ण \(\sqrt{10}\) बनता है। पिछले चरण को पहचानना जरूरी है।
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वर्गमूल सर्पिल में \(90^\circ\) कोण क्यों आवश्यक है?
Why is a \(90^\circ\) angle necessary in a square root spiral?
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A ताकि पाइथागोरस प्रमेय लगाया जा सके / So that Pythagoras theorem can be applied
B ताकि वृत्त बनाया जा सके / So that a circle can be made
C ताकि रेखा मिटाई जा सके / So that the line can be erased
D ताकि कोण \(0^\circ\) हो जाए / So that the angle becomes \(0^\circ\)
Explanation opens after your attempt
Correct Answer
A. ताकि पाइथागोरस प्रमेय लगाया जा सके / So that Pythagoras theorem can be applied
Step 1
Concept
Pythagoras theorem applies to a right triangle. Hence a \(90^\circ\) angle is necessary.
Step 2
Why this answer is correct
The correct answer is A. ताकि पाइथागोरस प्रमेय लगाया जा सके / So that Pythagoras theorem can be applied. Pythagoras theorem applies to a right triangle. Hence a \(90^\circ\) angle is necessary.
Step 3
Exam Tip
पाइथागोरस प्रमेय समकोण त्रिभुज में लागू होता है। इसलिए \(90^\circ\) कोण जरूरी है।
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यदि किसी चरण पर कर्ण \(\sqrt{n}\) है तो अगला कर्ण किस सूत्र से मिलेगा?
If the hypotenuse at a step is \(\sqrt{n}\), by which formula will the next hypotenuse be obtained?
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#square-root-spiral
#general-rule
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A \(\sqrt{n+1}\)
B \(\sqrt{n-1}\)
C \(\sqrt{2n}\)
D \(\sqrt{n+2}\)
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Correct Answer
A. \(\sqrt{n+1}\)
Step 1
Concept
The new hypotenuse is (\sqrt{\(\sqrt{n}\)2 +12 }=\sqrt{n+1}). This is the general rule of the 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 general rule of the spiral.
Step 3
Exam Tip
नया कर्ण (\sqrt{\(\sqrt{n}\)2 +12 }=\sqrt{n+1}) होता है। यही सर्पिल का सामान्य नियम है।
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वर्गमूल सर्पिल का उपयोग संख्या रेखा पर मुख्य रूप से किसे दिखाने में होता है?
A square root spiral is mainly used to show what on the number line?
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#square-root-spiral
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A वर्गमूल लंबाइयाँ / Square root lengths
B केवल ऋणात्मक पूर्णांक / Only negative integers
C केवल प्रतिशत / Only percentages
D केवल कोण / Only angles
Explanation opens after your attempt
Correct Answer
A. वर्गमूल लंबाइयाँ / Square root lengths
Step 1
Concept
A square root spiral gives lengths like \(\sqrt{2},\sqrt{3},\sqrt{5}\). These can be placed on the number line.
Step 2
Why this answer is correct
The correct answer is A. वर्गमूल लंबाइयाँ / Square root lengths. A square root spiral gives lengths like \(\sqrt{2},\sqrt{3},\sqrt{5}\). These can be placed on the number line.
Step 3
Exam Tip
वर्गमूल सर्पिल से \(\sqrt{2},\sqrt{3},\sqrt{5}\) जैसी लंबाइयाँ मिलती हैं। इन्हें संख्या रेखा पर रखा जा सकता है।
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वर्गमूल सर्पिल में \(\sqrt{5}\) की लंबाई किन दो पूर्ण संख्याओं के बीच होती है?
In a square root spiral, the length \(\sqrt{5}\) lies between which two whole numbers?
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#square-root-spiral
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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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वर्गमूल सर्पिल में \(\sqrt{12}\) बनाने से ठीक पहले कौन-सा कर्ण होना चाहिए?
Just before making \(\sqrt{12}\) in a square root spiral, which hypotenuse should be present?
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A \(\sqrt{10}\)
B \(\sqrt{11}\)
C \(\sqrt{12}\)
D \(\sqrt{13}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{11}\)
Step 1
Concept
Drawing a (1) unit perpendicular on \(\sqrt{11}\) gives \(\sqrt{12}\). The previous hypotenuse has one less number.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{11}\). Drawing a (1) unit perpendicular on \(\sqrt{11}\) gives \(\sqrt{12}\). The previous hypotenuse has one less number.
Step 3
Exam Tip
\(\sqrt{11}\) पर (1) इकाई लंब बनाने से \(\sqrt{12}\) मिलता है। पिछला कर्ण एक कम संख्या वाला होता है।
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वर्गमूल सर्पिल में \(\sqrt{15}\) के बाद कौन-सा कर्ण बनेगा?
In a square root spiral, which hypotenuse will be formed after \(\sqrt{15}\)?
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#next-root
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A \(\sqrt{14}\)
B \(\sqrt{15}\)
C \(\sqrt{16}\)
D \(\sqrt{30}\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{16}\)
Step 1
Concept
At each step the next hypotenuse is \(\sqrt{n+1}\). Therefore \(\sqrt{16}\) is formed after \(\sqrt{15}\).
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{16}\). At each step the next hypotenuse is \(\sqrt{n+1}\). Therefore \(\sqrt{16}\) is formed after \(\sqrt{15}\).
Step 3
Exam Tip
हर चरण में अगला कर्ण \(\sqrt{n+1}\) होता है। इसलिए \(\sqrt{15}\) के बाद \(\sqrt{16}\) बनेगा।
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वर्गमूल सर्पिल में \(\sqrt{16}\) किस पूर्ण संख्या के बराबर है?
In a square root spiral, \(\sqrt{16}\) is equal to which whole number?
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A (2)
B (3)
C (4)
D (8)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{16}=4\). Recognizing perfect squares helps in easy questions.
Step 2
Why this answer is correct
The correct answer is C. (4). \(\sqrt{16}=4\). Recognizing perfect squares helps in easy questions.
Step 3
Exam Tip
\(\sqrt{16}=4\) होता है। पूर्ण वर्ग पहचानना आसान प्रश्नों में मदद करता है।
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वर्गमूल सर्पिल में \(\sqrt{2}\) बनाने की सही गणना कौन-सी है?
Which is the correct calculation to make \(\sqrt{2}\) in a square root spiral?
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#sqrt2
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A \(\sqrt{1^2+1^2}\)
B \(\sqrt{2^2+2^2}\)
C \(\sqrt{1+2}\)
D \(\sqrt{2-1}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{1^2+1^2}\)
Step 1
Concept
Both sides of the first triangle are (1) unit. So the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{1^2+1^2}\). Both sides of the first triangle are (1) unit. So the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\).
Step 3
Exam Tip
पहले त्रिभुज की दोनों भुजाएँ (1) इकाई होती हैं। इसलिए कर्ण \(\sqrt{1^2+1^2}=\sqrt{2}\) है।
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वर्गमूल सर्पिल में \(\sqrt{3}\) बनाने की सही गणना कौन-सी है?
Which is the correct calculation to make \(\sqrt{3}\) in a square root spiral?
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A (\sqrt{\(\sqrt{2}\)2 +12 })
B \(\sqrt{2+2}\)
C \(\sqrt{3+1}\)
D \(\sqrt{1^2+3^2}\)
Explanation opens after your attempt
Correct Answer
A. (\sqrt{\(\sqrt{2}\)2 +12 })
Step 1
Concept
\(\sqrt{2}\) is the previous hypotenuse and (1) unit is the new perpendicular. Therefore the new hypotenuse is \(\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. (\sqrt{\(\sqrt{2}\)2 +12 }). \(\sqrt{2}\) is the previous hypotenuse and (1) unit is the new perpendicular. Therefore the new hypotenuse is \(\sqrt{3}\).
Step 3
Exam Tip
\(\sqrt{2}\) पिछले कर्ण की लंबाई है और (1) इकाई नई लंब है। इसलिए नया कर्ण \(\sqrt{3}\) मिलता है।
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वर्गमूल सर्पिल में (1) इकाई लंब भुजा बदलकर (2) इकाई कर दी जाए तो सामान्य क्रम क्यों बदल जाएगा?
If the (1) unit perpendicular side in a square root spiral is changed to (2) units, why will the usual sequence change?
#number-systems
#square-root-spiral
#concept
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A क्योंकि प्रत्येक चरण में (1) जोड़ने का नियम टूट जाएगा / Because the rule of adding (1) at each step will break
B क्योंकि कर्ण हमेशा \(\sqrt{1}\) रहेगा / Because the hypotenuse will always remain \(\sqrt{1}\)
C क्योंकि त्रिभुज नहीं बनेगा / Because no triangle will form
D क्योंकि कोण \(90^\circ\) नहीं हो सकता / Because the angle cannot be \(90^\circ\)
Explanation opens after your attempt
Correct Answer
A. क्योंकि प्रत्येक चरण में (1) जोड़ने का नियम टूट जाएगा / Because the rule of adding (1) at each step will break
Step 1
Concept
In the usual square root spiral \(1^2\) is added each time. Taking (2) units will change the sequence.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि प्रत्येक चरण में (1) जोड़ने का नियम टूट जाएगा / Because the rule of adding (1) at each step will break. In the usual square root spiral \(1^2\) is added each time. Taking (2) units will change the sequence.
Step 3
Exam Tip
सामान्य वर्गमूल सर्पिल में हर बार \(1^2\) जुड़ता है। (2) इकाई लेने पर क्रम अलग हो जाएगा।
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वर्गमूल सर्पिल में \(\sqrt{18}\) के बाद अगला कर्ण क्या होगा?
What will be the next hypotenuse after \(\sqrt{18}\) in a square root spiral?
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#square-root-spiral
#next-root
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A \(\sqrt{17}\)
B \(\sqrt{18}\)
C \(\sqrt{19}\)
D \(\sqrt{20}\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{19}\)
Step 1
Concept
At each new step a (1) unit perpendicular is added. So the number changes from (18) to (19).
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{19}\). At each new step a (1) unit perpendicular is added. So the number changes from (18) to (19).
Step 3
Exam Tip
हर नए चरण में (1) इकाई लंब जुड़ती है। इसलिए संख्या (18) से (19) हो जाती है।
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वर्गमूल सर्पिल में \(\sqrt{21}\) बनाने के लिए किस पिछले कर्ण का उपयोग होगा?
Which previous hypotenuse is used to make \(\sqrt{21}\) in a square root spiral?
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#previous-root
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A \(\sqrt{19}\)
B \(\sqrt{20}\)
C \(\sqrt{21}\)
D \(\sqrt{22}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{20}\)
Step 1
Concept
\(\sqrt{21}\) is formed from \(\sqrt{20}\) and a (1) unit perpendicular. The previous hypotenuse has one less number.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{20}\). \(\sqrt{21}\) is formed from \(\sqrt{20}\) and a (1) unit perpendicular. The previous hypotenuse has one less number.
Step 3
Exam Tip
\(\sqrt{20}\) और (1) इकाई लंब से \(\sqrt{21}\) बनता है। पिछले कर्ण में संख्या एक कम होती है।
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वर्गमूल सर्पिल में \(\sqrt{25}\) की लंबाई कितनी होगी?
What is the length of \(\sqrt{25}\) in a square root spiral?
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#perfect-square
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A (4)
B (5)
C (6)
D (25)
Explanation opens after your attempt
Step 1
Concept
\(\sqrt{25}=5\). It is the square root of a perfect square.
Step 2
Why this answer is correct
The correct answer is B. (5). \(\sqrt{25}=5\). It is the square root of a perfect square.
Step 3
Exam Tip
\(\sqrt{25}=5\) होता है। यह पूर्ण वर्ग का वर्गमूल है।
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वर्गमूल सर्पिल में कौन-सी लंबाई सामान्यतः अपरिमेय होगी?
Which length in a square root spiral is usually irrational?
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#irrational-number
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A \(\sqrt{1}\)
B \(\sqrt{4}\)
C \(\sqrt{8}\)
D \(\sqrt{9}\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{8}\)
Step 1
Concept
\(\sqrt{8}\) 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. \(\sqrt{8}\). \(\sqrt{8}\) is not the square root of a perfect square. Therefore it is irrational.
Step 3
Exam Tip
\(\sqrt{8}\) पूर्ण वर्ग का वर्गमूल नहीं है। इसलिए यह अपरिमेय है।
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वर्गमूल सर्पिल में \(\sqrt{4}\) के बाद बनने वाली अगली अपरिमेय लंबाई कौन-सी है?
In a square root spiral, which next irrational length is formed after \(\sqrt{4}\)?
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#irrational-length
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A \(\sqrt{5}\)
B \(\sqrt{4}\)
C \(\sqrt{9}\)
D \(\sqrt{16}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{5}\)
Step 1
Concept
\(\sqrt{4}=2\) is rational. After this \(\sqrt{5}\) is formed and it is irrational.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{5}\). \(\sqrt{4}=2\) is rational. After this \(\sqrt{5}\) is formed and it is irrational.
Step 3
Exam Tip
\(\sqrt{4}=2\) परिमेय है। इसके बाद \(\sqrt{5}\) बनता है जो अपरिमेय है।
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वर्गमूल सर्पिल में \(\sqrt{27}\) से अगला कर्ण कौन-सा बनेगा?
In a square root spiral, which hypotenuse is formed after \(\sqrt{27}\)?
#number-systems
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#next-root
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A \(\sqrt{26}\)
B \(\sqrt{27}\)
C \(\sqrt{28}\)
D \(\sqrt{29}\)
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Correct Answer
C. \(\sqrt{28}\)
Step 1
Concept
The next hypotenuse is \(\sqrt{27+1}=\sqrt{28}\). Remember the sequential rule.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{28}\). The next hypotenuse is \(\sqrt{27+1}=\sqrt{28}\). Remember the sequential rule.
Step 3
Exam Tip
अगला कर्ण \(\sqrt{27+1}=\sqrt{28}\) होगा। क्रमिक नियम याद रखें।
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वर्गमूल सर्पिल में \(\sqrt{30}\) बनाने से ठीक पहले कौन-सा कर्ण होगा?
Just before making \(\sqrt{30}\) in a square root spiral, which hypotenuse will be present?
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#previous-root
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A \(\sqrt{28}\)
B \(\sqrt{29}\)
C \(\sqrt{30}\)
D \(\sqrt{31}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{29}\)
Step 1
Concept
Drawing a (1) unit perpendicular on \(\sqrt{29}\) gives \(\sqrt{30}\). This is the construction order.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{29}\). Drawing a (1) unit perpendicular on \(\sqrt{29}\) gives \(\sqrt{30}\). This is the construction order.
Step 3
Exam Tip
\(\sqrt{29}\) पर (1) इकाई लंब बनाने से \(\sqrt{30}\) मिलता है। यही निर्माण का क्रम है।
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वर्गमूल सर्पिल में लंबाई को संख्या रेखा पर रखने के लिए किसकी सहायता ली जाती है?
What is used to place the length from a square root spiral on the number line?
#number-systems
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#compass
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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
Using a compass the hypotenuse length is transferred to the number line by drawing an arc. This gives the square root point.
Step 2
Why this answer is correct
The correct answer is A. कंपास / Compass. Using a compass the hypotenuse length is transferred to the number line by drawing an arc. This gives the square root point.
Step 3
Exam Tip
कंपास से कर्ण की लंबाई लेकर संख्या रेखा पर चाप लगाया जाता है। इससे वर्गमूल का बिंदु मिलता है।
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वर्गमूल सर्पिल में कौन-सा कथन सही है?
Which statement is correct for a square root spiral?
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A हर नया कर्ण पिछले से संबंधित अगला वर्गमूल देता है / Each new hypotenuse gives the next related square root
B हर नया कर्ण हमेशा (1) ही होता है / Each new hypotenuse is always (1)
C हर नया त्रिभुज समबाहु होता है / Each new triangle is equilateral
D हर नया कोण \(0^\circ\) होता है / Each new angle is \(0^\circ\)
Explanation opens after your attempt
Correct Answer
A. हर नया कर्ण पिछले से संबंधित अगला वर्गमूल देता है / Each new hypotenuse gives the next related square root
Step 1
Concept
In a square root spiral the new hypotenuse gives the next square root in order. It is made from right triangles.
Step 2
Why this answer is correct
The correct answer is A. हर नया कर्ण पिछले से संबंधित अगला वर्गमूल देता है / Each new hypotenuse gives the next related square root. In a square root spiral the new hypotenuse gives the next square root in order. It is made from right triangles.
Step 3
Exam Tip
वर्गमूल सर्पिल में नया कर्ण क्रम से अगला वर्गमूल देता है। यह समकोण त्रिभुजों से बनता है।
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वर्गमूल सर्पिल में \(\sqrt{2}\) को संख्या रेखा पर किसके बीच रखा जाता है?
In a square root spiral, \(\sqrt{2}\) is placed between which numbers on the number line?
#number-systems
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#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 \(1^2<2<2^2\). 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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वर्गमूल सर्पिल में \(\sqrt{10}\) किन पूर्ण संख्याओं के बीच होगा?
In a square root spiral, \(\sqrt{10}\) lies between which whole numbers?
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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 \(\sqrt{24}\) किन पूर्ण संख्याओं के बीच होगा?
In a square root spiral, \(\sqrt{24}\) lies between which whole numbers?
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A (3) और (4) / (3) and (4)
B (4) और (5) / (4) and (5)
C (5) और (6) / (5) and (6)
D (6) और (7) / (6) and (7)
Explanation opens after your attempt
Correct Answer
B. (4) और (5) / (4) and (5)
Step 1
Concept
Because \(4^2<24<5^2\). Therefore \(4<\sqrt{24}<5\).
Step 2
Why this answer is correct
The correct answer is B. (4) और (5) / (4) and (5). Because \(4^2<24<5^2\). Therefore \(4<\sqrt{24}<5\).
Step 3
Exam Tip
क्योंकि \(4^2<24<5^2\) है। इसलिए \(4<\sqrt{24}<5\) होता है।
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वर्गमूल सर्पिल में (1) इकाई लंब जोड़ने पर \(\sqrt{n}\) से \(\sqrt{n+1}\) क्यों बनता है?
Why does \(\sqrt{n}\) become \(\sqrt{n+1}\) after adding a (1) unit perpendicular in a square root spiral?
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#general-rule
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A क्योंकि (\(\sqrt{n}\)2 +12 =n+1) / Because (\(\sqrt{n}\)2 +12 =n+1)
B क्योंकि \(\sqrt{n}+1=\sqrt{n+1}\) / Because \(\sqrt{n}+1=\sqrt{n+1}\)
C क्योंकि (n-1=n+1) / Because (n-1=n+1)
D क्योंकि \(1^2=0\) / Because \(1^2=0\)
Explanation opens after your attempt
Correct Answer
A. क्योंकि (\(\sqrt{n}\)2 +12 =n+1) / Because (\(\sqrt{n}\)2 +12 =n+1)
Step 1
Concept
In Pythagoras theorem the squares of sides are added. Therefore the new hypotenuse is \(\sqrt{n+1}\).
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (\(\sqrt{n}\)2 +12 =n+1) / Because (\(\sqrt{n}\)2 +12 =n+1). In Pythagoras theorem the squares of sides are added. Therefore the new hypotenuse is \(\sqrt{n+1}\).
Step 3
Exam Tip
पाइथागोरस प्रमेय में भुजाओं के वर्ग जुड़ते हैं। इसलिए नया कर्ण \(\sqrt{n+1}\) होता है।
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वर्गमूल सर्पिल में \(\sqrt{3}\) से \(\sqrt{4}\) बनाने में कौन-सी बात गलत नहीं है?
In making \(\sqrt{4}\) from \(\sqrt{3}\) in a square root spiral, which statement is not wrong?
#number-systems
#square-root-spiral
#pythagoras
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A (\(\sqrt{3}\)2 +12 =4)
B \(\sqrt{3}+1=\sqrt{4}\)
C \(\sqrt{3}-1=\sqrt{4}\)
D (\(\sqrt{3}\)2 -12 =4)
Explanation opens after your attempt
Correct Answer
A. (\(\sqrt{3}\)2 +12 =4)
Step 1
Concept
The correct calculation uses the sum of squares. \(\sqrt{3}+1\) is not taken as \(\sqrt{4}\).
Step 2
Why this answer is correct
The correct answer is A. (\(\sqrt{3}\)2 +12 =4). The correct calculation uses the sum of squares. \(\sqrt{3}+1\) is not taken as \(\sqrt{4}\).
Step 3
Exam Tip
सही गणना वर्गों के योग से होती है। \(\sqrt{3}+1\) को \(\sqrt{4}\) नहीं माना जाता।
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वर्गमूल सर्पिल में \(\sqrt{32}\) बनाने के लिए पिछला कर्ण कौन-सा होगा?
To make \(\sqrt{32}\) in a square root spiral, which will be the previous hypotenuse?
#number-systems
#square-root-spiral
#previous-root
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A \(\sqrt{30}\)
B \(\sqrt{31}\)
C \(\sqrt{32}\)
D \(\sqrt{33}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{31}\)
Step 1
Concept
Adding a (1) unit perpendicular to \(\sqrt{31}\) gives \(\sqrt{32}\). The previous number is (1) less.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{31}\). Adding a (1) unit perpendicular to \(\sqrt{31}\) gives \(\sqrt{32}\). The previous number is (1) less.
Step 3
Exam Tip
\(\sqrt{31}\) में (1) इकाई लंब जोड़ने पर \(\sqrt{32}\) मिलता है। पिछली संख्या (1) कम होती है।
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वर्गमूल सर्पिल में \(\sqrt{36}\) की लंबाई क्या होगी?
What is the length of \(\sqrt{36}\) in a square root spiral?
#number-systems
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A (4)
B (5)
C (6)
D (7)
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 C. (6). \(\sqrt{36}=6\). The number (36) is a perfect square.
Step 3
Exam Tip
\(\sqrt{36}=6\) होता है। (36) एक पूर्ण वर्ग है।
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वर्गमूल सर्पिल में \(\sqrt{40}\) किस दो पूर्ण संख्याओं के बीच होगा?
In a square root spiral, \(\sqrt{40}\) lies between which two whole numbers?
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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 \(\sqrt{49}\) को देखकर कौन-सा निष्कर्ष सही है?
Which conclusion is correct by looking at \(\sqrt{49}\) in a square root spiral?
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A यह (7) के बराबर है / It is equal to (7)
B यह (6) के बराबर है / It is equal to (6)
C यह (49) के बराबर है / It is equal to (49)
D यह अपरिभाषित है / It is undefined
Explanation opens after your attempt
Correct Answer
A. यह (7) के बराबर है / It is equal to (7)
Step 1
Concept
\(\sqrt{49}=7\). Square roots of perfect squares are whole numbers.
Step 2
Why this answer is correct
The correct answer is A. यह (7) के बराबर है / It is equal to (7). \(\sqrt{49}=7\). Square roots of perfect squares are whole numbers.
Step 3
Exam Tip
\(\sqrt{49}=7\) होता है। पूर्ण वर्गों के वर्गमूल पूर्ण संख्या होते हैं।
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वर्गमूल सर्पिल में आकृति सर्पिल जैसी क्यों दिखती है?
Why does the figure in a square root spiral look like a spiral?
#number-systems
#square-root-spiral
#shape
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A क्योंकि समकोण त्रिभुज क्रम से घूमते हुए बनते हैं / Because right triangles are formed successively while turning
B क्योंकि सभी रेखाएँ समानांतर रहती हैं / Because all lines remain parallel
C क्योंकि कोई त्रिभुज नहीं बनता / Because no triangle is formed
D क्योंकि केवल वृत्त बनता है / Because only a circle is formed
Explanation opens after your attempt
Correct Answer
A. क्योंकि समकोण त्रिभुज क्रम से घूमते हुए बनते हैं / Because right triangles are formed successively while turning
Step 1
Concept
The direction of each new triangle changes. Continuous turning forms a spiral shape.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि समकोण त्रिभुज क्रम से घूमते हुए बनते हैं / Because right triangles are formed successively while turning. The direction of each new triangle changes. Continuous turning forms a spiral shape.
Step 3
Exam Tip
हर नए त्रिभुज की दिशा बदलती है। लगातार दिशा बदलने से सर्पिल आकृति बनती है।
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वर्गमूल सर्पिल में \(\sqrt{50}\) बनाने से पहले कौन-सा कर्ण बनेगा?
Before making \(\sqrt{50}\) in a square root spiral, which hypotenuse will be formed?
#number-systems
#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 \(\sqrt{50}\). Take the previous square root in order.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{49}\). Adding a (1) unit perpendicular to \(\sqrt{49}\) gives \(\sqrt{50}\). Take the previous square root in order.
Step 3
Exam Tip
\(\sqrt{49}\) पर (1) इकाई लंब जोड़ने से \(\sqrt{50}\) बनेगा। क्रम में पिछले वर्गमूल को लें।
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वर्गमूल सर्पिल में \(\sqrt{11}\) से नया कर्ण बनाने पर कौन-सा मिलेगा?
If a new hypotenuse is made from \(\sqrt{11}\) in a square root spiral, which one is obtained?
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#next-root
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A \(\sqrt{10}\)
B \(\sqrt{11}\)
C \(\sqrt{12}\)
D \(\sqrt{22}\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{12}\)
Step 1
Concept
Taking a (1) unit perpendicular with \(\sqrt{11}\) gives \(\sqrt{12}\). This is the step rule of the spiral.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{12}\). Taking a (1) unit perpendicular with \(\sqrt{11}\) gives \(\sqrt{12}\). This is the step rule of the spiral.
Step 3
Exam Tip
\(\sqrt{11}\) के साथ (1) इकाई लंब लेने पर \(\sqrt{12}\) बनता है। यह सर्पिल का क्रमिक नियम है।
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वर्गमूल सर्पिल में \(\sqrt{2}\) की लंबाई किस प्रकार की संख्या है?
In a square root spiral, what type of number is the length \(\sqrt{2}\)?
#number-systems
#square-root-spiral
#irrational-number
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A अपरिमेय संख्या / Irrational number
B पूर्ण संख्या / Whole number
C शून्य / Zero
D ऋणात्मक पूर्णांक / Negative integer
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
\(\sqrt{2}\) पूर्ण वर्ग का वर्गमूल नहीं है। इसलिए यह अपरिमेय संख्या है।
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वर्गमूल सर्पिल में \(\sqrt{3}\) का स्थान संख्या रेखा पर किसके बीच होगा?
In a square root spiral, the point for \(\sqrt{3}\) lies between which numbers on the 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 \(\sqrt{3}\) 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 \(\sqrt{17}\) किस दो पूर्ण संख्याओं के बीच होगा?
In a square root spiral, \(\sqrt{17}\) lies between which two whole numbers?
#number-systems
#square-root-spiral
#number-line
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A (3) और (4) / (3) and (4)
B (4) और (5) / (4) and (5)
C (5) और (6) / (5) and (6)
D (6) और (7) / (6) and (7)
Explanation opens after your attempt
Correct Answer
B. (4) और (5) / (4) and (5)
Step 1
Concept
Because \(4^2<17<5^2\). Therefore \(4<\sqrt{17}<5\).
Step 2
Why this answer is correct
The correct answer is B. (4) और (5) / (4) and (5). Because \(4^2<17<5^2\). Therefore \(4<\sqrt{17}<5\).
Step 3
Exam Tip
क्योंकि \(4^2<17<5^2\) है। इसलिए \(4<\sqrt{17}<5\) होता है।
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वर्गमूल सर्पिल में \(\sqrt{2}\) से \(\sqrt{3}\) बनने पर कौन-सी गलत धारणा से बचना चाहिए?
When \(\sqrt{3}\) is formed from \(\sqrt{2}\) in a square root spiral, which wrong idea should be avoided?
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A \(\sqrt{2}+1=\sqrt{3}\) मानना / Thinking \(\sqrt{2}+1=\sqrt{3}\)
B पाइथागोरस प्रमेय लगाना / Applying Pythagoras theorem
C (1) इकाई लंब जोड़ना / Adding (1) unit perpendicular
D समकोण त्रिभुज बनाना / Making a right triangle
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{2}+1=\sqrt{3}\) मानना / Thinking \(\sqrt{2}+1=\sqrt{3}\)
Step 1
Concept
\(\sqrt{3}\) is formed from the sum of squares, not direct addition. The correct calculation is (\(\sqrt{2}\)2 +12 =3).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}+1=\sqrt{3}\) मानना / Thinking \(\sqrt{2}+1=\sqrt{3}\). \(\sqrt{3}\) is formed from the sum of squares, not direct addition. The correct calculation is (\(\sqrt{2}\)2 +12 =3).
Step 3
Exam Tip
\(\sqrt{3}\) वर्गों के योग से बनता है, सीधे जोड़ से नहीं। सही गणना (\(\sqrt{2}\)2 +12 =3) है।
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वर्गमूल सर्पिल में कौन-सा उपकरण समकोण बनाने में सहायक हो सकता है?
Which tool can help in making a right angle in a square root spiral?
#number-systems
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#tools
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A सेट स्क्वायर / Set square
B तराजू / Balance
C घड़ी / Clock
D माप सिलेंडर / Measuring cylinder
Explanation opens after your attempt
Correct Answer
A. सेट स्क्वायर / Set square
Step 1
Concept
A set square helps make a \(90^\circ\) angle easily. The correct right angle gives the correct hypotenuse.
Step 2
Why this answer is correct
The correct answer is A. सेट स्क्वायर / Set square. A set square helps make a \(90^\circ\) angle easily. The correct right angle gives the correct hypotenuse.
Step 3
Exam Tip
सेट स्क्वायर से \(90^\circ\) कोण आसानी से बनाया जा सकता है। सही समकोण से ही सही कर्ण मिलता है।
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वर्गमूल सर्पिल में \(\sqrt{64}\) की लंबाई क्या होगी?
What is the length of \(\sqrt{64}\) in a square root spiral?
#number-systems
#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
\(\sqrt{64}=8\). The number (64) is a perfect square.
Step 2
Why this answer is correct
The correct answer is C. (8). \(\sqrt{64}=8\). The number (64) is a perfect square.
Step 3
Exam Tip
\(\sqrt{64}=8\) होता है। (64) पूर्ण वर्ग है।
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वर्गमूल सर्पिल में \(\sqrt{26}\) किस दो पूर्ण संख्याओं के बीच होगा?
In a square root spiral, \(\sqrt{26}\) lies between which two whole numbers?
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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 का मुख्य विचार एक वाक्य में क्या है?
What is the main idea of a square root spiral in one sentence?
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#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 निकालना / Finding only decimals
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
In a square root spiral each new right triangle gives the next square root. This is its main idea.
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. In a square root spiral each new right triangle gives the next square root. This is its main idea.
Step 3
Exam Tip
वर्गमूल सर्पिल में हर नया समकोण त्रिभुज अगला वर्गमूल देता है। यही इसका मुख्य विचार है।
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वर्गमूल सर्पिल से मिली लंबाई को संख्या रेखा पर अंकित करते समय कंपास की नोक सामान्यतः किस बिंदु पर रखी जाती है?
While marking the length obtained from a square root spiral on the number line, where is the compass point usually placed?
#number-systems
#square-root-spiral
#number-line
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A मूल बिंदु / Origin
B अंतिम त्रिभुज का लंब बिंदु / Perpendicular point of the last triangle
C कोई भी ऋणात्मक बिंदु / Any negative point
D कर्ण का मध्यबिंदु / Midpoint of the hypotenuse
Explanation opens after your attempt
Correct Answer
A. मूल बिंदु / Origin
Step 1
Concept
To show the length on the number line, the compass point is placed at the origin and an arc is drawn. This gives the correct square root point.
Step 2
Why this answer is correct
The correct answer is A. मूल बिंदु / Origin. To show the length on the number line, the compass point is placed at the origin and an arc is drawn. This gives the correct square root point.
Step 3
Exam Tip
लंबाई को संख्या रेखा पर दिखाने के लिए कंपास की नोक मूल बिंदु पर रखकर चाप खींचते हैं। इससे सही वर्गमूल बिंदु मिलता है।
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