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?
Explanation opens after your attempt
A. \(\sqrt{74}\) पर (1) इकाई लंब बनानी होगीDraw a (1) unit perpendicular on \(\sqrt{74}\)
Concept
\(\sqrt{75}\) is formed from \(\sqrt{74}\) and a (1) unit perpendicular. The previous hypotenuse always has one less number.
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.
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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