वर्गमूल सर्पिल में \(\sqrt{12}\) कौन से क्रम के समकोण त्रिभुज से प्राप्त होगा, यदि पहला त्रिभुज \(\sqrt{2}\) देता है?
In a square root spiral, \(\sqrt{12}\) is obtained from which numbered right triangle if the first triangle gives \(\sqrt{2}\)?
Explanation opens after your attempt
C. (11)वां(11)th
Concept
The first triangle gives \(\sqrt{2}\), so for \(\sqrt{12}\), the order is (12-1=11). Subtract (1) from the inner number.
Why this answer is correct
The correct answer is C. (11)वां / (11)th. The first triangle gives \(\sqrt{2}\), so for \(\sqrt{12}\), the order is (12-1=11). Subtract (1) from the inner number.
Exam Tip
पहला त्रिभुज \(\sqrt{2}\) देता है, इसलिए \(\sqrt{12}\) के लिए क्रम (12-1=11) होगा। अंदर की संख्या से (1) घटाएं।
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