Practice focused topic-wise MCQs with answers and explanations for quick revision and exam preparation.
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If \(OP=\sqrt{x}\) and the next hypotenuse is \(\sqrt{57}\), what is the value of (x)?
B. (56)
अगला कर्ण \(\sqrt{x+1}\) होता है, इसलिए (x+1=57)। इससे (x=56) मिलता है। / The next hypotenuse is \(\sqrt{x+1}\), so (x+1=57). This gives (x=56).
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Which statement is correct for the spiral from \(\sqrt{2}\) to \(\sqrt{9}\)?
A. इसमें (8) कर्ण बनते हैंIt has (8) hypotenuses
\(\sqrt{2}\) से \(\sqrt{9}\) तक (2,3,4,5,6,7,8,9) कुल (8) कर्ण हैं। \(\sqrt{4}\) और \(\sqrt{9}\) पूर्णांक लंबाइयां भी देते हैं। / From \(\sqrt{2}\) to \(\sqrt{9}\), the inner numbers (2,3,4,5,6,7,8,9) give (8) hypotenuses. \(\sqrt{4}\) and \(\sqrt{9}\) also give integer lengths.
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If \(\sqrt{27}\) is to be simplified as a length of the spiral, what is the correct form?
B. \(3\sqrt{3}\)
\(\sqrt{27}=\sqrt{9\times3}=3\sqrt{3}\)। मूल लंबाई वही है, केवल रूप सरल किया गया है। / \(\sqrt{27}=\sqrt{9\times3}=3\sqrt{3}\). The length is the same, only the form is simplified.
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If \(OP=\sqrt{18}\) and the new (PQ=1) is perpendicular, what will be the simplest form of (OQ)?
C. \(\sqrt{19}\)
नया कर्ण \(\sqrt{18+1}=\sqrt{19}\) होगा। (19) पूर्ण वर्ग नहीं है, इसलिए यही सरल रूप है। / The new hypotenuse is \(\sqrt{18+1}=\sqrt{19}\). Since (19) is not a perfect square, this is already the simplest form.
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Which option explains why \(\sqrt{4}\) is not irrational even though it is formed in the spiral?
A. क्योंकि \(\sqrt{4}=2\)Because \(\sqrt{4}=2\)
\(\sqrt{4}=2\) है, जो परिमेय संख्या है। सर्पिल में बने सभी वर्गमूल अपरिमेय नहीं होते। / \(\sqrt{4}=2\), which is a rational number. Not all square roots formed in the spiral are irrational.
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If the hypotenuse of the (m)th right triangle is \(\sqrt{46}\), what is the value of (m)?
B. (45)
(m)वें त्रिभुज का कर्ण \(\sqrt{m+1}\) होता है। इसलिए (m+1=46) से (m=45)। / The hypotenuse of the (m)th triangle is \(\sqrt{m+1}\). Thus (m+1=46), so (m=45).
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B. (3)
\(\sqrt{4}=2\), \(\sqrt{9}=3\), और \(\sqrt{16}=4\) पूर्णांक हैं। इसलिए कुल (3) पूर्णांक लंबाइयां मिलेंगी। / \(\sqrt{4}=2\), \(\sqrt{9}=3\), and \(\sqrt{16}=4\) are integers. Therefore, (3) integer lengths are obtained.
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Which option gives the correct general recurrence for the square root spiral?
A. \(अगला कर्ण (=\sqrt{\)पिछले कर्ण का वर्ग+1})\(Next hypotenuse (=\sqrt{\)square of previous hypotenuse+1})
\(पाइथागोरस प्रमेय से अगला कर्ण (\sqrt{(\)पिछला कर्ण)2+1}) होता है। इसलिए वर्ग और वर्गमूल दोनों सही जगह होने चाहिए। \(/ By Pythagoras theorem, the next hypotenuse is (\sqrt{(\)previous hypotenuse\()^2+1}). So both square and square root must be in the correct places.\)
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If the value of a hypotenuse is (5) units, as which square root will it appear in the spiral?
C. \(\sqrt{25}\)
लंबाई (5) होने पर वर्ग (25) होगा, इसलिए कर्ण \(\sqrt{25}\) के रूप में दिखेगा। पूर्णांक लंबाइयों को उनके वर्गमूल रूप से जोड़ें। / If the length is (5), its square is (25), so the hypotenuse appears as \(\sqrt{25}\). Connect integer lengths with their square root forms.
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What is the most important mathematical basis of the square root spiral?
B. पाइथागोरस प्रमेयPythagoras theorem
वर्गमूल सर्पिल में हर नया कर्ण समकोण त्रिभुज से बनता है। इसलिए इसका मुख्य आधार पाइथागोरस प्रमेय है। / In a square root spiral, every new hypotenuse is formed using a right triangle. Therefore, its main basis is Pythagoras theorem.
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