वर्गमूल सर्पिल में \(\sqrt{3}\) से \(\sqrt{4}\) बनाने में कौन-सी बात गलत नहीं है?

In making \(\sqrt{4}\) from \(\sqrt{3}\) in a square root spiral, which statement is not wrong?

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

A. (\(\sqrt{3}\)2+12=4)

Step 1

Concept

The correct calculation uses the sum of squares. \(\sqrt{3}+1\) is not taken as \(\sqrt{4}\).

Step 2

Why this answer is correct

The correct answer is A. (\(\sqrt{3}\)2+12=4). The correct calculation uses the sum of squares. \(\sqrt{3}+1\) is not taken as \(\sqrt{4}\).

Step 3

Exam Tip

सही गणना वर्गों के योग से होती है। \(\sqrt{3}+1\) को \(\sqrt{4}\) नहीं माना जाता।

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Mathematics Answer, Explanation and Revision Hints

वर्गमूल सर्पिल में \(\sqrt{3}\) से \(\sqrt{4}\) बनाने में कौन-सी बात गलत नहीं है? / In making \(\sqrt{4}\) from \(\sqrt{3}\) in a square root spiral, which statement is not wrong?

Correct Answer: A. (\(\sqrt{3}\)2+12=4). Explanation: सही गणना वर्गों के योग से होती है। \(\sqrt{3}+1\) को \(\sqrt{4}\) नहीं माना जाता। / The correct calculation uses the sum of squares. \(\sqrt{3}+1\) is not taken as \(\sqrt{4}\).

Which concept should I revise for this Mathematics MCQ?

The correct calculation uses the sum of squares. \(\sqrt{3}+1\) is not taken as \(\sqrt{4}\).

What exam hint can help solve this Mathematics question?

सही गणना वर्गों के योग से होती है। \(\sqrt{3}+1\) को \(\sqrt{4}\) नहीं माना जाता।