वर्गमूल सर्पिल में \(\sqrt{2}\) बनाने की सही गणना कौन-सी है?
Which is the correct calculation to make \(\sqrt{2}\) in a square root spiral?
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{1^2+1^2}\)
Concept
Both sides of the first triangle are (1) unit. So the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\).
Why this answer is correct
The correct answer is A. \(\sqrt{1^2+1^2}\). Both sides of the first triangle are (1) unit. So the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\).
Exam Tip
पहले त्रिभुज की दोनों भुजाएँ (1) इकाई होती हैं। इसलिए कर्ण \(\sqrt{1^2+1^2}=\sqrt{2}\) है।
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