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In a square root spiral, the new hypotenuse is \(\sqrt{n+1}\). If the previous hypotenuse was \(\sqrt{1368}\), what will be the new hypotenuse?वर्गमूल सर्पिल में नया कर्ण \(\sqrt{n+1}\) होता है। यदि पिछला कर्ण \(\sqrt{1368}\) था, तो नया कर्ण क्या होगा?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(\sqrt{1369}\)\(\sqrt{1369}\)
Previous hypotenuse is \(\sqrt{1368}=\sqrt{n}\). Given that the new hypotenuse is \(\sqrt{n+1}\), we get \(n+1=1368+1=1369\). Hence the new hypotenuse is \(\sqrt{1369}=37\).पिछला कर्ण \(\sqrt{1368}=\sqrt{n}\) है। दिया है कि नया कर्ण \(\sqrt{n+1}\) होता है, इसलिए \(n+1=1368+1=1369\)। अतः नया कर्ण \(\sqrt{1369}=37\) होगा।
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