वर्गमूल सर्पिल में (1) इकाई लंब जोड़ने पर \(\sqrt{n}\) से \(\sqrt{n+1}\) क्यों बनता है?

Why does \(\sqrt{n}\) become \(\sqrt{n+1}\) after adding a (1) unit perpendicular in a square root spiral?

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

A. क्योंकि (\(\sqrt{n}\)2+12=n+1)Because (\(\sqrt{n}\)2+12=n+1)

Step 1

Concept

In Pythagoras theorem the squares of sides are added. Therefore the new hypotenuse is \(\sqrt{n+1}\).

Step 2

Why this answer is correct

The correct answer is A. क्योंकि (\(\sqrt{n}\)2+12=n+1) / Because (\(\sqrt{n}\)2+12=n+1). In Pythagoras theorem the squares of sides are added. Therefore the new hypotenuse is \(\sqrt{n+1}\).

Step 3

Exam Tip

पाइथागोरस प्रमेय में भुजाओं के वर्ग जुड़ते हैं। इसलिए नया कर्ण \(\sqrt{n+1}\) होता है।

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वर्गमूल सर्पिल में (1) इकाई लंब जोड़ने पर \(\sqrt{n}\) से \(\sqrt{n+1}\) क्यों बनता है? / Why does \(\sqrt{n}\) become \(\sqrt{n+1}\) after adding a (1) unit perpendicular in a square root spiral?

Correct Answer: A. क्योंकि (\(\sqrt{n}\)2+12=n+1) / Because (\(\sqrt{n}\)2+12=n+1). Explanation: पाइथागोरस प्रमेय में भुजाओं के वर्ग जुड़ते हैं। इसलिए नया कर्ण \(\sqrt{n+1}\) होता है। / In Pythagoras theorem the squares of sides are added. Therefore the new hypotenuse is \(\sqrt{n+1}\).

Which concept should I revise for this Mathematics MCQ?

In Pythagoras theorem the squares of sides are added. Therefore the new hypotenuse is \(\sqrt{n+1}\).

What exam hint can help solve this Mathematics question?

पाइथागोरस प्रमेय में भुजाओं के वर्ग जुड़ते हैं। इसलिए नया कर्ण \(\sqrt{n+1}\) होता है।