वर्गमूल सर्पिल में यदि पिछला कर्ण \(\sqrt{13}\) है, तो (1) इकाई लंब जोड़ने पर नए कर्ण की सही गणना कौन-सी होगी?

In a square root spiral, if the previous hypotenuse is \(\sqrt{13}\), which calculation correctly gives the new hypotenuse after adding a (1) unit perpendicular?

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

A. \(\sqrt{13+1}\)

Step 1

Concept

By Pythagoras the new hypotenuse is (\sqrt{\(\sqrt{13}\)2+12}=\sqrt{14}). In a square root spiral the number increases by one.

Step 2

Why this answer is correct

The correct answer is A. \(\sqrt{13+1}\). By Pythagoras the new hypotenuse is (\sqrt{\(\sqrt{13}\)2+12}=\sqrt{14}). In a square root spiral the number increases by one.

Step 3

Exam Tip

पाइथागोरस से नया कर्ण (\sqrt{\(\sqrt{13}\)2+12}=\sqrt{14}) होगा। वर्गमूल सर्पिल में संख्या एक बढ़ती है।

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वर्गमूल सर्पिल में यदि पिछला कर्ण \(\sqrt{13}\) है, तो (1) इकाई लंब जोड़ने पर नए कर्ण की सही गणना कौन-सी होगी? / In a square root spiral, if the previous hypotenuse is \(\sqrt{13}\), which calculation correctly gives the new hypotenuse after adding a (1) unit perpendicular?

Correct Answer: A. \(\sqrt{13+1}\). Explanation: पाइथागोरस से नया कर्ण (\sqrt{\(\sqrt{13}\)2+12}=\sqrt{14}) होगा। वर्गमूल सर्पिल में संख्या एक बढ़ती है। / By Pythagoras the new hypotenuse is (\sqrt{\(\sqrt{13}\)2+12}=\sqrt{14}). In a square root spiral the number increases by one.

Which concept should I revise for this Mathematics MCQ?

By Pythagoras the new hypotenuse is (\sqrt{\(\sqrt{13}\)2+12}=\sqrt{14}). In a square root spiral the number increases by one.

What exam hint can help solve this Mathematics question?

पाइथागोरस से नया कर्ण (\sqrt{\(\sqrt{13}\)2+12}=\sqrt{14}) होगा। वर्गमूल सर्पिल में संख्या एक बढ़ती है।