वर्गमूल सर्पिल में \(\sqrt{32}\) बनाने के लिए पिछला कर्ण कौन-सा होगा?
To make \(\sqrt{32}\) in a square root spiral, which will be the previous hypotenuse?
Explanation opens after your attempt
B. \(\sqrt{31}\)
Concept
Adding a (1) unit perpendicular to \(\sqrt{31}\) gives \(\sqrt{32}\). The previous number is (1) less.
Why this answer is correct
The correct answer is B. \(\sqrt{31}\). Adding a (1) unit perpendicular to \(\sqrt{31}\) gives \(\sqrt{32}\). The previous number is (1) less.
Exam Tip
\(\sqrt{31}\) में (1) इकाई लंब जोड़ने पर \(\sqrt{32}\) मिलता है। पिछली संख्या (1) कम होती है।
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