वर्गमूल सर्पिल में यदि \(\sqrt{80}\) कर्ण पर (1) इकाई लंब बनाई जाए, तो नया कर्ण किस अंतराल में होगा?

If a (1) unit perpendicular is drawn on hypotenuse \(\sqrt{80}\) in a square root spiral, in which interval will the new hypotenuse lie?

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

C. ठीक (9) परExactly at (9)

Step 1

Concept

The new hypotenuse will be \(\sqrt{81}\), and \(\sqrt{81}=9\). Therefore it lies exactly at (9).

Step 2

Why this answer is correct

The correct answer is C. ठीक (9) पर / Exactly at (9). The new hypotenuse will be \(\sqrt{81}\), and \(\sqrt{81}=9\). Therefore it lies exactly at (9).

Step 3

Exam Tip

नया कर्ण \(\sqrt{81}\) होगा और \(\sqrt{81}=9\) है। इसलिए वह ठीक (9) पर स्थित होगा।

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वर्गमूल सर्पिल में यदि \(\sqrt{80}\) कर्ण पर (1) इकाई लंब बनाई जाए, तो नया कर्ण किस अंतराल में होगा? / If a (1) unit perpendicular is drawn on hypotenuse \(\sqrt{80}\) in a square root spiral, in which interval will the new hypotenuse lie?

Correct Answer: C. ठीक (9) पर / Exactly at (9). Explanation: नया कर्ण \(\sqrt{81}\) होगा और \(\sqrt{81}=9\) है। इसलिए वह ठीक (9) पर स्थित होगा। / The new hypotenuse will be \(\sqrt{81}\), and \(\sqrt{81}=9\). Therefore it lies exactly at (9).

Which concept should I revise for this Mathematics MCQ?

The new hypotenuse will be \(\sqrt{81}\), and \(\sqrt{81}=9\). Therefore it lies exactly at (9).

What exam hint can help solve this Mathematics question?

नया कर्ण \(\sqrt{81}\) होगा और \(\sqrt{81}=9\) है। इसलिए वह ठीक (9) पर स्थित होगा।