Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
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If \(OP=\sqrt{15}\), (PQ=1), and \(PQ \perp OP\), what is the value of \(OQ^2\)?
C. (16)
(OQ-2=\(\sqrt{15}\)2+12=16) होगा। प्रश्न में \(OQ^2\) पूछा है, इसलिए उत्तर (16) है। / (OQ-2=\(\sqrt{15}\)2+12=16). Since the question asks for \(OQ^2\), the answer is (16).
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In a square root spiral, the point representing \(\sqrt{5}\) comes immediately after which one?
B. \(\sqrt{4}\)
क्रम \(\sqrt{2},\sqrt{3},\sqrt{4},\sqrt{5}\) चलता है। इसलिए \(\sqrt{5}\), \(\sqrt{4}\) के तुरंत बाद आता है। / The order is \(\sqrt{2},\sqrt{3},\sqrt{4},\sqrt{5}\). Therefore, \(\sqrt{5}\) comes immediately after \(\sqrt{4}\).
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Why are segments like \(\sqrt{2},\sqrt{3},\sqrt{4}\) formed in a square root spiral?
B. समकोण त्रिभुज और पाइथागोरस प्रमेय का प्रयोग होता हैRight triangles and Pythagoras theorem are used
सर्पिल की पूरी रचना समकोण त्रिभुजों पर आधारित है। पाइथागोरस प्रमेय से नए कर्ण की लंबाई मिलती है। / The whole construction of the spiral is based on right triangles. Pythagoras theorem gives the length of the new hypotenuse.
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B. (5)
पहला त्रिभुज \(\sqrt{2}\) देता है और पांचवां त्रिभुज \(\sqrt{6}\) देता है। इसलिए कुल (5) समकोण त्रिभुज बनते हैं। / The first triangle gives \(\sqrt{2}\), and the fifth triangle gives \(\sqrt{6}\). So, (5) right triangles are formed.
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If the hypotenuse of the fourth constructed right triangle is asked, what is the correct length?
B. \(\sqrt{5}\)
पहला त्रिभुज \(\sqrt{2}\), दूसरा \(\sqrt{3}\), तीसरा \(\sqrt{4}\), चौथा \(\sqrt{5}\) देता है। चरण संख्या में (1) जोड़ें। / The first triangle gives \(\sqrt{2}\), the second \(\sqrt{3}\), the third \(\sqrt{4}\), and the fourth \(\sqrt{5}\). Add (1) to the step number.
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B. \(\sqrt{12}\)
(\(\sqrt{12}\)2+12=13), इसलिए \(\sqrt{12}\) वाले कर्ण पर अगला लंब बनेगा। पिछले कर्ण को एक कम अंदर संख्या से पहचानें। / (\(\sqrt{12}\)2+12=13), so the next perpendicular is drawn on the hypotenuse \(\sqrt{12}\). Identify the previous hypotenuse by reducing the inner number by (1).
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C. \(\sqrt{18}\)
अगले चरण में (1) इकाई लंब जुड़ता है, इसलिए नई दूरी \(\sqrt{17+1}=\sqrt{18}\) होगी। वर्गमूल के अंदर की संख्या बढ़ती है। / In the next step, a perpendicular of (1) unit is added, so the new distance becomes \(\sqrt{17+1}=\sqrt{18}\). The number under the root increases.
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Which pair correctly shows two consecutive hypotenuses of the square root spiral?
B. \(\sqrt{9},\sqrt{10}\)
लगातार कर्णों में वर्गमूल के अंदर की संख्याएं लगातार होती हैं। इसलिए \(\sqrt{9}\) और \(\sqrt{10}\) सही युग्म है। / For consecutive hypotenuses, the numbers under the roots are consecutive. Therefore, \(\sqrt{9}\) and \(\sqrt{10}\) form the correct pair.
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If \(OP=\sqrt{20}\), what was the length of the hypotenuse formed just before it?
B. \(\sqrt{19}\)
ठीक पहले वाले चरण में वर्गमूल के अंदर की संख्या (1) कम होगी, इसलिए लंबाई \(\sqrt{19}\) है। क्रम को पीछे से पढ़ना भी सीखें। / In the immediately previous step, the number under the root is (1) less, so the length is \(\sqrt{19}\). Learn to read the sequence backward too.
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Looking at \(\sqrt{4}\) and \(\sqrt{9}\) in a square root spiral, which conclusion is correct?
B. दोनों परिमेय संख्याएं हैंBoth are rational numbers
\(\sqrt{4}=2\) और \(\sqrt{9}=3\), इसलिए दोनों परिमेय हैं। सर्पिल परिमेय और अपरिमेय दोनों प्रकार के वर्गमूल दिखा सकता है। / \(\sqrt{4}=2\) and \(\sqrt{9}=3\), so both are rational. The spiral can show both rational and irrational square roots.
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