यदि किसी चरण पर कर्ण \(\sqrt{n}\) है तो अगला कर्ण किस सूत्र से मिलेगा?

If the hypotenuse at a step is \(\sqrt{n}\), by which formula will the next hypotenuse be obtained?

Author: Muft Shiksha Editorial Team Published:
Explanation opens after your attempt
Correct Answer

A. \(\sqrt{n+1}\)

Step 1

Concept

The new hypotenuse is (\sqrt{\(\sqrt{n}\)2+12}=\sqrt{n+1}). This is the general rule of the spiral.

Step 2

Why this answer is correct

The correct answer is A. \(\sqrt{n+1}\). The new hypotenuse is (\sqrt{\(\sqrt{n}\)2+12}=\sqrt{n+1}). This is the general rule of the spiral.

Step 3

Exam Tip

नया कर्ण (\sqrt{\(\sqrt{n}\)2+12}=\sqrt{n+1}) होता है। यही सर्पिल का सामान्य नियम है।

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यदि किसी चरण पर कर्ण \(\sqrt{n}\) है तो अगला कर्ण किस सूत्र से मिलेगा? / If the hypotenuse at a step is \(\sqrt{n}\), by which formula will the next hypotenuse be obtained?

Correct Answer: A. \(\sqrt{n+1}\). Explanation: नया कर्ण (\sqrt{\(\sqrt{n}\)2+12}=\sqrt{n+1}) होता है। यही सर्पिल का सामान्य नियम है। / The new hypotenuse is (\sqrt{\(\sqrt{n}\)2+12}=\sqrt{n+1}). This is the general rule of the spiral.

Which concept should I revise for this Mathematics MCQ?

The new hypotenuse is (\sqrt{\(\sqrt{n}\)2+12}=\sqrt{n+1}). This is the general rule of the spiral.

What exam hint can help solve this Mathematics question?

नया कर्ण (\sqrt{\(\sqrt{n}\)2+12}=\sqrt{n+1}) होता है। यही सर्पिल का सामान्य नियम है।