यदि कोई सर्पिल \(\sqrt{1}\) से शुरू होकर \(\sqrt{n}\) तक जाता है, तो बनाए गए समकोण त्रिभुजों की संख्या क्या होगी?
If a spiral starts from \(\sqrt{1}\) and goes up to \(\sqrt{n}\), how many right triangles are constructed?
Explanation opens after your attempt
Correct Answer
B. (n-1)
Concept
\(\sqrt{2}\) is given by the first triangle, so up to \(\sqrt{n}\), the number of triangles is (n-1). Remember the starting term while writing a general formula.
Why this answer is correct
The correct answer is B. (n-1). \(\sqrt{2}\) is given by the first triangle, so up to \(\sqrt{n}\), the number of triangles is (n-1). Remember the starting term while writing a general formula.
Exam Tip
\(\sqrt{2}\) पहला त्रिभुज देता है, इसलिए \(\sqrt{n}\) तक त्रिभुजों की संख्या (n-1) होगी। सामान्य सूत्र लिखते समय आरंभिक पद याद रखें।
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