वर्गमूल सर्पिल में \(\sqrt{80}\) बनाने के लिए पिछला कर्ण कौन-सा होगा?
To construct \(\sqrt{80}\) in a square root spiral, which previous hypotenuse will be used?
Explanation opens after your attempt
B. \(\sqrt{79}\)
Concept
Adding a (1) unit perpendicular to \(\sqrt{79}\) gives \(\sqrt{80}\). The previous hypotenuse has one less number.
Why this answer is correct
The correct answer is B. \(\sqrt{79}\). Adding a (1) unit perpendicular to \(\sqrt{79}\) gives \(\sqrt{80}\). The previous hypotenuse has one less number.
Exam Tip
\(\sqrt{79}\) पर (1) इकाई लंब जोड़ने से \(\sqrt{80}\) बनता है। पिछला कर्ण एक कम संख्या का होता है।
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