वर्गमूल सर्पिल में यदि \(\sqrt{9603}\) कर्ण पर (1) इकाई लंब बनाई जाए, तो नया कर्ण कौन-सा होगा और उसका सटीक मान क्या होगा?
If a (1) unit perpendicular is drawn on hypotenuse \(\sqrt{9603}\) in a square root spiral, what will be the new hypotenuse and its exact value?
Explanation opens after your attempt
A. \(\sqrt{9604}=98\)
Concept
The new hypotenuse is \(\sqrt{9603+1}=\sqrt{9604}\). Since \(9604=98^2\), the exact value is (98).
Why this answer is correct
The correct answer is A. \(\sqrt{9604}=98\). The new hypotenuse is \(\sqrt{9603+1}=\sqrt{9604}\). Since \(9604=98^2\), the exact value is (98).
Exam Tip
नया कर्ण \(\sqrt{9603+1}=\sqrt{9604}\) होगा। \(9604=98^2\), इसलिए सटीक मान (98) है।
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