वर्गमूल सर्पिल में \(\sqrt{59}\) बनाने के लिए कौन-सा पिछला कर्ण सही है और नया कर्ण किस अंतराल में होगा?
To construct \(\sqrt{59}\) in a square root spiral, which previous hypotenuse is correct and in which interval will the new hypotenuse lie?
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{58}\), (7) और (8) के बीच\(\sqrt{58}\), between (7) and (8)
Concept
\(\sqrt{59}\) is formed from \(\sqrt{58}\), and \(7^2<59<8^2\). Therefore it lies between (7) and (8).
Why this answer is correct
The correct answer is A. \(\sqrt{58}\), (7) और (8) के बीच / \(\sqrt{58}\), between (7) and (8). \(\sqrt{59}\) is formed from \(\sqrt{58}\), and \(7^2<59<8^2\). Therefore it lies between (7) and (8).
Exam Tip
\(\sqrt{58}\) से \(\sqrt{59}\) बनता है और \(7^2<59<8^2\) है। इसलिए यह (7) और (8) के बीच है।
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