Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
Difficulty select karke sirf is topic ke focused MCQs practice karein. Score, timer aur explanations student-friendly flow me milenge.
Showing 121-130 of 650 questions.
B. \(\sqrt{17}\)
\(\sqrt{17}\) में (1) इकाई लंब जोड़कर \(\sqrt{18}\) मिलता है। पिछले कर्ण की संख्या एक कम होती है। / Adding a (1) unit perpendicular to \(\sqrt{17}\) gives \(\sqrt{18}\). The previous hypotenuse has one less number.
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In a square root spiral, which hypotenuse comes after \(\sqrt{30}\)?
C. \(\sqrt{31}\)
अगले चरण में संख्या (1) बढ़ती है। इसलिए \(\sqrt{30}\) के बाद \(\sqrt{31}\) आता है। / In the next step the number increases by (1). Therefore \(\sqrt{31}\) comes after \(\sqrt{30}\).
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In a square root spiral, \(\sqrt{36}\) will be equal to what?
B. (6)
\(\sqrt{36}=6\) होता है। (36) पूर्ण वर्ग है। / \(\sqrt{36}=6\). The number (36) is a perfect square.
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Between which whole numbers will \(\sqrt{26}\) lie on the number line?
B. (5) और (6)(5) and (6)
क्योंकि \(5^2<26<6^2\) है। इसलिए \(5<\sqrt{26}<6\) होता है। / Because \(5^2<26<6^2\). Therefore \(5<\sqrt{26}<6\).
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Why is the (1) unit perpendicular drawn at \(90^\circ\) in a square root spiral?
A. समकोण त्रिभुज बनाने के लिएTo make a right triangle
\(90^\circ\) से समकोण त्रिभुज बनता है। फिर पाइथागोरस प्रमेय से नया कर्ण मिलता है। / A \(90^\circ\) angle makes a right triangle. Then Pythagoras theorem gives the new hypotenuse.
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Which previous hypotenuse is used to construct \(\sqrt{50}\) in a square root spiral?
B. \(\sqrt{49}\)
\(\sqrt{49}\) में (1) इकाई लंब जोड़ने पर \(\sqrt{50}\) बनता है। पिछली संख्या (1) कम होती है। / Adding a (1) unit perpendicular to \(\sqrt{49}\) gives \(\sqrt{50}\). The previous number is (1) less.
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What is the length of \(\sqrt{49}\) in a square root spiral?
B. (7)
\(\sqrt{49}=7\) होता है। (49) पूर्ण वर्ग है। / \(\sqrt{49}=7\). The number (49) is a perfect square.
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Between which whole numbers will \(\sqrt{40}\) lie on the number line?
B. (6) और (7)(6) and (7)
क्योंकि \(6^2<40<7^2\) है। इसलिए \(6<\sqrt{40}<7\) होता है। / Because \(6^2<40<7^2\). Therefore \(6<\sqrt{40}<7\).
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If a new hypotenuse is made from \(\sqrt{6}\) in a square root spiral, which hypotenuse is obtained?
C. \(\sqrt{7}\)
\(\sqrt{6}\) में (1) इकाई लंब जोड़ने से \(\sqrt{7}\) बनता है। यह पाइथागोरस प्रमेय से आता है। / Adding a (1) unit perpendicular to \(\sqrt{6}\) forms \(\sqrt{7}\). This follows from Pythagoras theorem.
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In a square root spiral, \(\sqrt{64}\) is equal to which number?
C. (8)
\(\sqrt{64}=8\) होता है। पूर्ण वर्गों की पहचान से उत्तर जल्दी मिलता है। / \(\sqrt{64}=8\). Recognizing perfect squares gives the answer quickly.
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