वर्गमूल सर्पिल में \(\sqrt{143}\) पर (1) इकाई लंब बनाने से कौन-सा कर्ण बनेगा और उसका मान कहाँ होगा?

In a square root spiral, drawing a (1) unit perpendicular on \(\sqrt{143}\) forms which hypotenuse and where will its value lie?

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

D. \(\sqrt{144}\), ठीक (12) पर\(\sqrt{144}\), exactly at (12)

Step 1

Concept

The new hypotenuse is \(\sqrt{144}\). Since \(\sqrt{144}=12\), it is not in an interval but exactly at (12).

Step 2

Why this answer is correct

The correct answer is D. \(\sqrt{144}\), ठीक (12) पर / \(\sqrt{144}\), exactly at (12). The new hypotenuse is \(\sqrt{144}\). Since \(\sqrt{144}=12\), it is not in an interval but exactly at (12).

Step 3

Exam Tip

नया कर्ण \(\sqrt{144}\) है। क्योंकि \(\sqrt{144}=12\), यह किसी अंतराल में नहीं बल्कि ठीक (12) पर 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 \(\sqrt{143}\) पर (1) इकाई लंब बनाने से कौन-सा कर्ण बनेगा और उसका मान कहाँ होगा? / In a square root spiral, drawing a (1) unit perpendicular on \(\sqrt{143}\) forms which hypotenuse and where will its value lie?

Correct Answer: D. \(\sqrt{144}\), ठीक (12) पर / \(\sqrt{144}\), exactly at (12). Explanation: नया कर्ण \(\sqrt{144}\) है। क्योंकि \(\sqrt{144}=12\), यह किसी अंतराल में नहीं बल्कि ठीक (12) पर है। / The new hypotenuse is \(\sqrt{144}\). Since \(\sqrt{144}=12\), it is not in an interval but exactly at (12).

Which concept should I revise for this Mathematics MCQ?

The new hypotenuse is \(\sqrt{144}\). Since \(\sqrt{144}=12\), it is not in an interval but exactly at (12).

What exam hint can help solve this Mathematics question?

नया कर्ण \(\sqrt{144}\) है। क्योंकि \(\sqrt{144}=12\), यह किसी अंतराल में नहीं बल्कि ठीक (12) पर है।