वर्गमूल सर्पिल में \(\sqrt{75}\) बनाने के लिए पिछले कर्ण पर क्या करना होगा?

What must be done on the previous hypotenuse to construct \(\sqrt{75}\) in a square root spiral?

Author: Muft Shiksha Editorial Team Published:
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Correct Answer

A. \(\sqrt{74}\) पर (1) इकाई लंब बनानी होगीDraw a (1) unit perpendicular on \(\sqrt{74}\)

Step 1

Concept

\(\sqrt{75}\) is formed from \(\sqrt{74}\) and a (1) unit perpendicular. The previous hypotenuse always has one less number.

Step 2

Why this answer is correct

The correct answer is A. \(\sqrt{74}\) पर (1) इकाई लंब बनानी होगी / Draw a (1) unit perpendicular on \(\sqrt{74}\). \(\sqrt{75}\) is formed from \(\sqrt{74}\) and a (1) unit perpendicular. The previous hypotenuse always has one less number.

Step 3

Exam Tip

\(\sqrt{74}\) और (1) इकाई लंब से \(\sqrt{75}\) बनता है। पिछला कर्ण हमेशा एक कम संख्या वाला होता है।

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वर्गमूल सर्पिल में \(\sqrt{75}\) बनाने के लिए पिछले कर्ण पर क्या करना होगा? / What must be done on the previous hypotenuse to construct \(\sqrt{75}\) in a square root spiral?

Correct Answer: A. \(\sqrt{74}\) पर (1) इकाई लंब बनानी होगी / Draw a (1) unit perpendicular on \(\sqrt{74}\). Explanation: \(\sqrt{74}\) और (1) इकाई लंब से \(\sqrt{75}\) बनता है। पिछला कर्ण हमेशा एक कम संख्या वाला होता है। / \(\sqrt{75}\) is formed from \(\sqrt{74}\) and a (1) unit perpendicular. The previous hypotenuse always has one less number.

Which concept should I revise for this Mathematics MCQ?

\(\sqrt{75}\) is formed from \(\sqrt{74}\) and a (1) unit perpendicular. The previous hypotenuse always has one less number.

What exam hint can help solve this Mathematics question?

\(\sqrt{74}\) और (1) इकाई लंब से \(\sqrt{75}\) बनता है। पिछला कर्ण हमेशा एक कम संख्या वाला होता है।