वर्गमूल सर्पिल में \(\sqrt{7}\) को दर्शाने वाला खंड किस प्रकार मिलता है?

How is the segment representing \(\sqrt{7}\) obtained in a square root spiral?

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

B. \(\sqrt{6}\) पर (1) इकाई लंब बनाकरBy drawing (1) unit perpendicular on \(\sqrt{6}\)

Step 1

Concept

Because (\(\sqrt{6}\)2+12=7), the new segment becomes \(\sqrt{7}\). Identifying the correct previous segment is necessary.

Step 2

Why this answer is correct

The correct answer is B. \(\sqrt{6}\) पर (1) इकाई लंब बनाकर / By drawing (1) unit perpendicular on \(\sqrt{6}\). Because (\(\sqrt{6}\)2+12=7), the new segment becomes \(\sqrt{7}\). Identifying the correct previous segment is necessary.

Step 3

Exam Tip

क्योंकि (\(\sqrt{6}\)2+12=7), इसलिए नया खंड \(\sqrt{7}\) होगा। सही पिछले खंड को पहचानना जरूरी 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 \(\sqrt{7}\) को दर्शाने वाला खंड किस प्रकार मिलता है? / How is the segment representing \(\sqrt{7}\) obtained in a square root spiral?

Correct Answer: B. \(\sqrt{6}\) पर (1) इकाई लंब बनाकर / By drawing (1) unit perpendicular on \(\sqrt{6}\). Explanation: क्योंकि (\(\sqrt{6}\)2+12=7), इसलिए नया खंड \(\sqrt{7}\) होगा। सही पिछले खंड को पहचानना जरूरी है। / Because (\(\sqrt{6}\)2+12=7), the new segment becomes \(\sqrt{7}\). Identifying the correct previous segment is necessary.

Which concept should I revise for this Mathematics MCQ?

Because (\(\sqrt{6}\)2+12=7), the new segment becomes \(\sqrt{7}\). Identifying the correct previous segment is necessary.

What exam hint can help solve this Mathematics question?

क्योंकि (\(\sqrt{6}\)2+12=7), इसलिए नया खंड \(\sqrt{7}\) होगा। सही पिछले खंड को पहचानना जरूरी है।