वर्गमूल सर्पिल में \(\sqrt{12}\) बनाने से ठीक पहले कौन-सा कर्ण होना चाहिए?

Just before making \(\sqrt{12}\) in a square root spiral, which hypotenuse should be present?

Author: Muft Shiksha Editorial Team Published:
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Correct Answer

B. \(\sqrt{11}\)

Step 1

Concept

Drawing a (1) unit perpendicular on \(\sqrt{11}\) gives \(\sqrt{12}\). The previous hypotenuse has one less number.

Step 2

Why this answer is correct

The correct answer is B. \(\sqrt{11}\). Drawing a (1) unit perpendicular on \(\sqrt{11}\) gives \(\sqrt{12}\). The previous hypotenuse has one less number.

Step 3

Exam Tip

\(\sqrt{11}\) पर (1) इकाई लंब बनाने से \(\sqrt{12}\) मिलता है। पिछला कर्ण एक कम संख्या वाला होता है।

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वर्गमूल सर्पिल में \(\sqrt{12}\) बनाने से ठीक पहले कौन-सा कर्ण होना चाहिए? / Just before making \(\sqrt{12}\) in a square root spiral, which hypotenuse should be present?

Correct Answer: B. \(\sqrt{11}\). Explanation: \(\sqrt{11}\) पर (1) इकाई लंब बनाने से \(\sqrt{12}\) मिलता है। पिछला कर्ण एक कम संख्या वाला होता है। / Drawing a (1) unit perpendicular on \(\sqrt{11}\) gives \(\sqrt{12}\). The previous hypotenuse has one less number.

Which concept should I revise for this Mathematics MCQ?

Drawing a (1) unit perpendicular on \(\sqrt{11}\) gives \(\sqrt{12}\). The previous hypotenuse has one less number.

What exam hint can help solve this Mathematics question?

\(\sqrt{11}\) पर (1) इकाई लंब बनाने से \(\sqrt{12}\) मिलता है। पिछला कर्ण एक कम संख्या वाला होता है।