किस विकल्प में वर्गमूल सर्पिल की लंबाइयों का सही आरंभिक क्रम है?
Which option shows the correct initial sequence of lengths in a square root spiral?
Explanation opens after your attempt
Correct Answer
A. \(1,\sqrt{2},\sqrt{3},\sqrt{4}\)
Concept
The start is (OA=1), and then the hypotenuses \(\sqrt{2},\sqrt{3},\sqrt{4}\) are formed. The number under the root increases continuously.
Why this answer is correct
The correct answer is A. \(1,\sqrt{2},\sqrt{3},\sqrt{4}\). The start is (OA=1), and then the hypotenuses \(\sqrt{2},\sqrt{3},\sqrt{4}\) are formed. The number under the root increases continuously.
Exam Tip
आरंभ (OA=1) से होता है और फिर कर्ण \(\sqrt{2},\sqrt{3},\sqrt{4}\) बनते हैं। क्रम में अंदर की संख्या लगातार बढ़ती है।
Login to save your score, XP, coins and progress.