वर्गमूल सर्पिल में यदि \(\sqrt{80}\) कर्ण पर (1) इकाई लंब बनाई जाए, तो नया कर्ण किस अंतराल में होगा?
If a (1) unit perpendicular is drawn on hypotenuse \(\sqrt{80}\) in a square root spiral, in which interval will the new hypotenuse lie?
Explanation opens after your attempt
C. ठीक (9) परExactly at (9)
Concept
The new hypotenuse will be \(\sqrt{81}\), and \(\sqrt{81}=9\). Therefore it lies exactly at (9).
Why this answer is correct
The correct answer is C. ठीक (9) पर / Exactly at (9). The new hypotenuse will be \(\sqrt{81}\), and \(\sqrt{81}=9\). Therefore it lies exactly at (9).
Exam Tip
नया कर्ण \(\sqrt{81}\) होगा और \(\sqrt{81}=9\) है। इसलिए वह ठीक (9) पर स्थित होगा।
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