वर्गमूल सर्पिल में \(\sqrt{6}\) से नया कर्ण बनाने पर कौन-सा कर्ण मिलेगा?

If a new hypotenuse is made from \(\sqrt{6}\) in a square root spiral, which hypotenuse is obtained?

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

C. \(\sqrt{7}\)

Step 1

Concept

Adding a (1) unit perpendicular to \(\sqrt{6}\) forms \(\sqrt{7}\). This follows from Pythagoras theorem.

Step 2

Why this answer is correct

The correct answer is C. \(\sqrt{7}\). Adding a (1) unit perpendicular to \(\sqrt{6}\) forms \(\sqrt{7}\). This follows from Pythagoras theorem.

Step 3

Exam Tip

\(\sqrt{6}\) में (1) इकाई लंब जोड़ने से \(\sqrt{7}\) बनता है। यह पाइथागोरस प्रमेय से आता है।

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Mathematics Answer, Explanation and Revision Hints

वर्गमूल सर्पिल में \(\sqrt{6}\) से नया कर्ण बनाने पर कौन-सा कर्ण मिलेगा? / If a new hypotenuse is made from \(\sqrt{6}\) in a square root spiral, which hypotenuse is obtained?

Correct Answer: C. \(\sqrt{7}\). Explanation: \(\sqrt{6}\) में (1) इकाई लंब जोड़ने से \(\sqrt{7}\) बनता है। यह पाइथागोरस प्रमेय से आता है। / Adding a (1) unit perpendicular to \(\sqrt{6}\) forms \(\sqrt{7}\). This follows from Pythagoras theorem.

Which concept should I revise for this Mathematics MCQ?

Adding a (1) unit perpendicular to \(\sqrt{6}\) forms \(\sqrt{7}\). This follows from Pythagoras theorem.

What exam hint can help solve this Mathematics question?

\(\sqrt{6}\) में (1) इकाई लंब जोड़ने से \(\sqrt{7}\) बनता है। यह पाइथागोरस प्रमेय से आता है।