वर्गमूल सर्पिल में \(\sqrt{18}\) बनाने के लिए किस पिछले कर्ण पर (1) इकाई लंब बनाई जाएगी?

To make \(\sqrt{18}\) in a square root spiral, on which previous hypotenuse is a (1) unit perpendicular drawn?

Author: Muft Shiksha Editorial Team Published:
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Correct Answer

B. \(\sqrt{17}\)

Step 1

Concept

With \(\sqrt{17}\) and a (1) unit perpendicular, \(\sqrt{18}\) is formed. The previous number is always (1) less.

Step 2

Why this answer is correct

The correct answer is B. \(\sqrt{17}\). With \(\sqrt{17}\) and a (1) unit perpendicular, \(\sqrt{18}\) is formed. The previous number is always (1) less.

Step 3

Exam Tip

\(\sqrt{17}\) के साथ (1) इकाई लंब से \(\sqrt{18}\) बनता है। पिछली संख्या हमेशा (1) कम होती है।

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Mathematics Answer, Explanation and Revision Hints

वर्गमूल सर्पिल में \(\sqrt{18}\) बनाने के लिए किस पिछले कर्ण पर (1) इकाई लंब बनाई जाएगी? / To make \(\sqrt{18}\) in a square root spiral, on which previous hypotenuse is a (1) unit perpendicular drawn?

Correct Answer: B. \(\sqrt{17}\). Explanation: \(\sqrt{17}\) के साथ (1) इकाई लंब से \(\sqrt{18}\) बनता है। पिछली संख्या हमेशा (1) कम होती है। / With \(\sqrt{17}\) and a (1) unit perpendicular, \(\sqrt{18}\) is formed. The previous number is always (1) less.

Which concept should I revise for this Mathematics MCQ?

With \(\sqrt{17}\) and a (1) unit perpendicular, \(\sqrt{18}\) is formed. The previous number is always (1) less.

What exam hint can help solve this Mathematics question?

\(\sqrt{17}\) के साथ (1) इकाई लंब से \(\sqrt{18}\) बनता है। पिछली संख्या हमेशा (1) कम होती है।