वर्गमूल सर्पिल में \(\sqrt{169}\) 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 \(\sqrt{169}\) gives which new hypotenuse and where is it located?

Author: Muft Shiksha Editorial Team Published:
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Correct Answer

A. \(\sqrt{170}\), (13) और (14) के बीच\(\sqrt{170}\), between (13) and (14)

Step 1

Concept

The new hypotenuse is \(\sqrt{170}\). Since \(13^2<170<14^2\), it lies between (13) and (14).

Step 2

Why this answer is correct

The correct answer is A. \(\sqrt{170}\), (13) और (14) के बीच / \(\sqrt{170}\), between (13) and (14). The new hypotenuse is \(\sqrt{170}\). Since \(13^2<170<14^2\), it lies between (13) and (14).

Step 3

Exam Tip

नया कर्ण \(\sqrt{170}\) है। क्योंकि \(13^2<170<14^2\), यह (13) और (14) के बीच है।

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वर्गमूल सर्पिल में \(\sqrt{169}\) 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 \(\sqrt{169}\) gives which new hypotenuse and where is it located?

Correct Answer: A. \(\sqrt{170}\), (13) और (14) के बीच / \(\sqrt{170}\), between (13) and (14). Explanation: नया कर्ण \(\sqrt{170}\) है। क्योंकि \(13^2<170<14^2\), यह (13) और (14) के बीच है। / The new hypotenuse is \(\sqrt{170}\). Since \(13^2<170<14^2\), it lies between (13) and (14).

Which concept should I revise for this Mathematics MCQ?

The new hypotenuse is \(\sqrt{170}\). Since \(13^2<170<14^2\), it lies between (13) and (14).

What exam hint can help solve this Mathematics question?

नया कर्ण \(\sqrt{170}\) है। क्योंकि \(13^2<170<14^2\), यह (13) और (14) के बीच है।