वर्गमूल सर्पिल में \(\sqrt{12}\) बनाने से ठीक पहले कौन-सा कर्ण होना चाहिए?
Just before making \(\sqrt{12}\) in a square root spiral, which hypotenuse should be present?
Explanation opens after your attempt
B. \(\sqrt{11}\)
Concept
Drawing a (1) unit perpendicular on \(\sqrt{11}\) gives \(\sqrt{12}\). The previous hypotenuse has one less number.
Why this answer is correct
The correct answer is B. \(\sqrt{11}\). Drawing a (1) unit perpendicular on \(\sqrt{11}\) gives \(\sqrt{12}\). The previous hypotenuse has one less number.
Exam Tip
\(\sqrt{11}\) पर (1) इकाई लंब बनाने से \(\sqrt{12}\) मिलता है। पिछला कर्ण एक कम संख्या वाला होता है।
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