वर्गमूल सर्पिल में \(\sqrt{32}\) बनाने के लिए पिछला कर्ण कौन-सा होगा?

To make \(\sqrt{32}\) in a square root spiral, which will be the previous hypotenuse?

Author: Muft Shiksha Editorial Team Published:
Explanation opens after your attempt
Correct Answer

B. \(\sqrt{31}\)

Step 1

Concept

Adding a (1) unit perpendicular to \(\sqrt{31}\) gives \(\sqrt{32}\). The previous number is (1) less.

Step 2

Why this answer is correct

The correct answer is B. \(\sqrt{31}\). Adding a (1) unit perpendicular to \(\sqrt{31}\) gives \(\sqrt{32}\). The previous number is (1) less.

Step 3

Exam Tip

\(\sqrt{31}\) में (1) इकाई लंब जोड़ने पर \(\sqrt{32}\) मिलता है। पिछली संख्या (1) कम होती है।

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Mathematics Answer, Explanation and Revision Hints

वर्गमूल सर्पिल में \(\sqrt{32}\) बनाने के लिए पिछला कर्ण कौन-सा होगा? / To make \(\sqrt{32}\) in a square root spiral, which will be the previous hypotenuse?

Correct Answer: B. \(\sqrt{31}\). Explanation: \(\sqrt{31}\) में (1) इकाई लंब जोड़ने पर \(\sqrt{32}\) मिलता है। पिछली संख्या (1) कम होती है। / Adding a (1) unit perpendicular to \(\sqrt{31}\) gives \(\sqrt{32}\). The previous number is (1) less.

Which concept should I revise for this Mathematics MCQ?

Adding a (1) unit perpendicular to \(\sqrt{31}\) gives \(\sqrt{32}\). The previous number is (1) less.

What exam hint can help solve this Mathematics question?

\(\sqrt{31}\) में (1) इकाई लंब जोड़ने पर \(\sqrt{32}\) मिलता है। पिछली संख्या (1) कम होती है।