Question:
Find the third proportional of \((3 + \sqrt{2})\) and \(2\sqrt{7}\).
\((3 + \sqrt{2})\) और \(2\sqrt{7}\) का तीसरा समानुपाती (third proportional) ज्ञात कीजिए।
Options:
- (A) \(2(3 + \sqrt{2})\)
- (B) \((12 – \sqrt{8})\)
- (C) \(4(3 – \sqrt{2})\)
- (D) \((12 + \sqrt{8})\)
- (A) \(2(3 + \sqrt{2})\)
- (B) \((12 – \sqrt{8})\)
- (C) \(4(3 – \sqrt{2})\)
- (D) \((12 + \sqrt{8})\)
Correct Answer: Option C
Explanation:
Detailed Explanation: Third proportional \(x = \frac{b^2}{a} = \frac{(2\sqrt{7})^2}{3 + \sqrt{2}} = \frac{28}{3 + \sqrt{2}} = \frac{28(3 – \sqrt{2})}{9 – 2} = \mathbf{4(3 – \sqrt{2})}\).
विस्तृत व्याख्या: तीसरा समानुपाती \(x = \frac{b^2}{a} = \frac{(2\sqrt{7})^2}{3 + \sqrt{2}} = \frac{28}{3 + \sqrt{2}}\)
परिमेयकरण करने पर: \(\frac{28(3 – \sqrt{2})}{(3)^2 – (\sqrt{2})^2} = \frac{28(3 – \sqrt{2})}{9 – 2} = \frac{28(3 – \sqrt{2})}{7} = \mathbf{4(3 – \sqrt{2})}\)।