वर्गमूल सर्पिल में \(\sqrt{143}\) पर (1) इकाई लंब बनाने से कौन-सा कर्ण बनेगा और उसका मान कहाँ होगा?
In a square root spiral, drawing a (1) unit perpendicular on \(\sqrt{143}\) forms which hypotenuse and where will its value lie?
Explanation opens after your attempt
D. \(\sqrt{144}\), ठीक (12) पर\(\sqrt{144}\), exactly at (12)
Concept
The new hypotenuse is \(\sqrt{144}\). Since \(\sqrt{144}=12\), it is not in an interval but exactly at (12).
Why this answer is correct
The correct answer is D. \(\sqrt{144}\), ठीक (12) पर / \(\sqrt{144}\), exactly at (12). The new hypotenuse is \(\sqrt{144}\). Since \(\sqrt{144}=12\), it is not in an interval but exactly at (12).
Exam Tip
नया कर्ण \(\sqrt{144}\) है। क्योंकि \(\sqrt{144}=12\), यह किसी अंतराल में नहीं बल्कि ठीक (12) पर है।
Login to save your score, XP, coins and progress.
