यदि \(r=\sqrt{28}+\sqrt{112}\) है, तो (r) किसके बराबर है?

If \(r=\sqrt{28}+\sqrt{112}\), what is (r) equal to?

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

A. \(6\sqrt{7}\)

Step 1

Concept

\(\sqrt{28}=2\sqrt{7}\) and \(\sqrt{112}=4\sqrt{7}\). Therefore \(r=6\sqrt{7}\).

Step 2

Why this answer is correct

The correct answer is A. \(6\sqrt{7}\). \(\sqrt{28}=2\sqrt{7}\) and \(\sqrt{112}=4\sqrt{7}\). Therefore \(r=6\sqrt{7}\).

Step 3

Exam Tip

\(\sqrt{28}=2\sqrt{7}\) और \(\sqrt{112}=4\sqrt{7}\) है। इसलिए \(r=6\sqrt{7}\) है।

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Mathematics Answer, Explanation and Revision Hints

यदि \(r=\sqrt{28}+\sqrt{112}\) है, तो (r) किसके बराबर है? / If \(r=\sqrt{28}+\sqrt{112}\), what is (r) equal to?

Correct Answer: A. \(6\sqrt{7}\). Explanation: \(\sqrt{28}=2\sqrt{7}\) और \(\sqrt{112}=4\sqrt{7}\) है। इसलिए \(r=6\sqrt{7}\) है। / \(\sqrt{28}=2\sqrt{7}\) and \(\sqrt{112}=4\sqrt{7}\). Therefore \(r=6\sqrt{7}\).

Which concept should I revise for this Mathematics MCQ?

\(\sqrt{28}=2\sqrt{7}\) and \(\sqrt{112}=4\sqrt{7}\). Therefore \(r=6\sqrt{7}\).

What exam hint can help solve this Mathematics question?

\(\sqrt{28}=2\sqrt{7}\) और \(\sqrt{112}=4\sqrt{7}\) है। इसलिए \(r=6\sqrt{7}\) है।