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Just before making (\sqrt{12}) in a square root spiral, which hypotenuse should be present?वर्गमूल सर्पिल में (\sqrt{12}) बनाने से ठीक पहले कौन-सा कर्ण होना चाहिए?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: (\sqrt{11})(\sqrt{11})
Drawing a (1) unit perpendicular on (\sqrt{11}) gives (\sqrt{12}). The previous hypotenuse has one less number.(\sqrt{11}) पर (1) इकाई लंब बनाने से (\sqrt{12}) मिलता है। पिछला कर्ण एक कम संख्या वाला होता है।
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