AP EAMCET · Maths · Properties of Triangles
If the lengths of two sides of a triangle are the roots of the equation \(x^2-2 \sqrt{3} x+2=0\) and the angle between these sides is \(\frac{\pi}{3}\), then the perimeter of the triangle is
- A \(2 \sqrt{6}+\sqrt{3}\)
- B \(2 \sqrt{6}+2 \sqrt{3}\)
- C \(\sqrt{6}+2 \sqrt{3}\)
- D \(\sqrt{6}+\sqrt{3}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{6}+2 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
\(x^2 - 2\sqrt{3}x + 2 = 0\) \(a+b = 2\sqrt{3}\) \(ab = 2\) \(c^2 = a^2+b^2-2ab \cos(\frac{\pi}{3})\) \(c^2 = (a+b)^2-2ab-2ab(\frac{1}{2})\) \(c^2 = (2\sqrt{3})^2-2(2)-2(2)(\frac{1}{2})\) \(c^2 = 12-4-2 = 6\) \(c = \sqrt{6}\) Perimeter \(P = a+b+c\) \(P = 2\sqrt{3} + \sqrt{6}\)
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