वर्गमूल सर्पिल में \(\sqrt{624}\) पर (1) इकाई लंब बनाने से नया कर्ण कौन-सा होगा और कहाँ स्थित होगा?

In a square root spiral, drawing a (1) unit perpendicular on \(\sqrt{624}\) gives which new hypotenuse and where is it located?

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

A. \(\sqrt{625}\), ठीक (25) पर\(\sqrt{625}\), exactly at (25)

Step 1

Concept

The new hypotenuse is \(\sqrt{625}\). Since \(\sqrt{625}=25\), it will be located exactly at (25).

Step 2

Why this answer is correct

The correct answer is A. \(\sqrt{625}\), ठीक (25) पर / \(\sqrt{625}\), exactly at (25). The new hypotenuse is \(\sqrt{625}\). Since \(\sqrt{625}=25\), it will be located exactly at (25).

Step 3

Exam Tip

नया कर्ण \(\sqrt{625}\) है। क्योंकि \(\sqrt{625}=25\), यह ठीक (25) पर स्थित होगा।

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वर्गमूल सर्पिल में \(\sqrt{624}\) पर (1) इकाई लंब बनाने से नया कर्ण कौन-सा होगा और कहाँ स्थित होगा? / In a square root spiral, drawing a (1) unit perpendicular on \(\sqrt{624}\) gives which new hypotenuse and where is it located?

Correct Answer: A. \(\sqrt{625}\), ठीक (25) पर / \(\sqrt{625}\), exactly at (25). Explanation: नया कर्ण \(\sqrt{625}\) है। क्योंकि \(\sqrt{625}=25\), यह ठीक (25) पर स्थित होगा। / The new hypotenuse is \(\sqrt{625}\). Since \(\sqrt{625}=25\), it will be located exactly at (25).

Which concept should I revise for this Mathematics MCQ?

The new hypotenuse is \(\sqrt{625}\). Since \(\sqrt{625}=25\), it will be located exactly at (25).

What exam hint can help solve this Mathematics question?

नया कर्ण \(\sqrt{625}\) है। क्योंकि \(\sqrt{625}=25\), यह ठीक (25) पर स्थित होगा।