वर्गमूल सर्पिल में \(\sqrt{21}\) बनाने के लिए कौन-सा पिछला कर्ण और नई लंब भुजा सही है?

To construct \(\sqrt{21}\) in a square root spiral, which previous hypotenuse and new perpendicular side are correct?

Author: Muft Shiksha Editorial Team Published:
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Correct Answer

B. \(\sqrt{20}\) और (1)\(\sqrt{20}\) and (1)

Step 1

Concept

Since (\(\sqrt{20}\)2+12=21). Therefore \(\sqrt{20}\) is the correct previous hypotenuse.

Step 2

Why this answer is correct

The correct answer is B. \(\sqrt{20}\) और (1) / \(\sqrt{20}\) and (1). Since (\(\sqrt{20}\)2+12=21). Therefore \(\sqrt{20}\) is the correct previous hypotenuse.

Step 3

Exam Tip

(\(\sqrt{20}\)2+12=21) होता है। इसलिए \(\sqrt{20}\) पिछले कर्ण के रूप में सही है।

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वर्गमूल सर्पिल में \(\sqrt{21}\) बनाने के लिए कौन-सा पिछला कर्ण और नई लंब भुजा सही है? / To construct \(\sqrt{21}\) in a square root spiral, which previous hypotenuse and new perpendicular side are correct?

Correct Answer: B. \(\sqrt{20}\) और (1) / \(\sqrt{20}\) and (1). Explanation: (\(\sqrt{20}\)2+12=21) होता है। इसलिए \(\sqrt{20}\) पिछले कर्ण के रूप में सही है। / Since (\(\sqrt{20}\)2+12=21). Therefore \(\sqrt{20}\) is the correct previous hypotenuse.

Which concept should I revise for this Mathematics MCQ?

Since (\(\sqrt{20}\)2+12=21). Therefore \(\sqrt{20}\) is the correct previous hypotenuse.

What exam hint can help solve this Mathematics question?

(\(\sqrt{20}\)2+12=21) होता है। इसलिए \(\sqrt{20}\) पिछले कर्ण के रूप में सही है।