वर्गमूल सर्पिल में \(\sqrt{7}\) बनाने से ठीक पहले कौन-सा कर्ण बनता है?
Which hypotenuse is constructed just before \(\sqrt{7}\) in a square root spiral?
Explanation opens after your attempt
B. \(\sqrt{6}\)
Concept
Adding a (1) unit perpendicular to \(\sqrt{6}\) gives \(\sqrt{7}\). The previous hypotenuse has one less number.
Why this answer is correct
The correct answer is B. \(\sqrt{6}\). Adding a (1) unit perpendicular to \(\sqrt{6}\) gives \(\sqrt{7}\). The previous hypotenuse has one less number.
Exam Tip
\(\sqrt{6}\) में (1) इकाई लंब जोड़ने पर \(\sqrt{7}\) मिलता है। पिछला कर्ण हमेशा एक कम संख्या का होता है।
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