वर्गमूल सर्पिल में \(\sqrt{80}\) बनाने के लिए पिछला कर्ण कौन-सा होगा?

To construct \(\sqrt{80}\) in a square root spiral, which previous hypotenuse will be used?

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

B. \(\sqrt{79}\)

Step 1

Concept

Adding a (1) unit perpendicular to \(\sqrt{79}\) gives \(\sqrt{80}\). The previous hypotenuse has one less number.

Step 2

Why this answer is correct

The correct answer is B. \(\sqrt{79}\). Adding a (1) unit perpendicular to \(\sqrt{79}\) gives \(\sqrt{80}\). The previous hypotenuse has one less number.

Step 3

Exam Tip

\(\sqrt{79}\) पर (1) इकाई लंब जोड़ने से \(\sqrt{80}\) बनता है। पिछला कर्ण एक कम संख्या का होता है।

Question me issue ya doubt hai?

Answer, explanation, typing mistake ya suggestion directly hamari team ko bhejein. 📱Helpline (Call / WhatsApp): +91 7272824365

Related Mathematics Questions

FAQs

Mathematics Answer, Explanation and Revision Hints

वर्गमूल सर्पिल में \(\sqrt{80}\) बनाने के लिए पिछला कर्ण कौन-सा होगा? / To construct \(\sqrt{80}\) in a square root spiral, which previous hypotenuse will be used?

Correct Answer: B. \(\sqrt{79}\). Explanation: \(\sqrt{79}\) पर (1) इकाई लंब जोड़ने से \(\sqrt{80}\) बनता है। पिछला कर्ण एक कम संख्या का होता है। / Adding a (1) unit perpendicular to \(\sqrt{79}\) gives \(\sqrt{80}\). The previous hypotenuse has one less number.

Which concept should I revise for this Mathematics MCQ?

Adding a (1) unit perpendicular to \(\sqrt{79}\) gives \(\sqrt{80}\). The previous hypotenuse has one less number.

What exam hint can help solve this Mathematics question?

\(\sqrt{79}\) पर (1) इकाई लंब जोड़ने से \(\sqrt{80}\) बनता है। पिछला कर्ण एक कम संख्या का होता है।