वर्गमूल सर्पिल में \(\sqrt{15}\) बनाने से ठीक पहले कौन-सा कर्ण होगा?

Just before constructing \(\sqrt{15}\) in a square root spiral, which hypotenuse will be present?

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

B. \(\sqrt{14}\)

Step 1

Concept

With \(\sqrt{14}\) and a (1) unit perpendicular, \(\sqrt{15}\) is formed. The previous hypotenuse has one less number.

Step 2

Why this answer is correct

The correct answer is B. \(\sqrt{14}\). With \(\sqrt{14}\) and a (1) unit perpendicular, \(\sqrt{15}\) is formed. The previous hypotenuse has one less number.

Step 3

Exam Tip

\(\sqrt{14}\) के साथ (1) इकाई लंब से \(\sqrt{15}\) बनता है। पिछला कर्ण एक कम संख्या का 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 \(\sqrt{15}\) बनाने से ठीक पहले कौन-सा कर्ण होगा? / Just before constructing \(\sqrt{15}\) in a square root spiral, which hypotenuse will be present?

Correct Answer: B. \(\sqrt{14}\). Explanation: \(\sqrt{14}\) के साथ (1) इकाई लंब से \(\sqrt{15}\) बनता है। पिछला कर्ण एक कम संख्या का है। / With \(\sqrt{14}\) and a (1) unit perpendicular, \(\sqrt{15}\) is formed. The previous hypotenuse has one less number.

Which concept should I revise for this Mathematics MCQ?

With \(\sqrt{14}\) and a (1) unit perpendicular, \(\sqrt{15}\) is formed. The previous hypotenuse has one less number.

What exam hint can help solve this Mathematics question?

\(\sqrt{14}\) के साथ (1) इकाई लंब से \(\sqrt{15}\) बनता है। पिछला कर्ण एक कम संख्या का है।