Two sides of a triangle are given by the roots of the equation& x^ 2 - 2 3 x + 2 = 0 and the angle between the sides is (π)/(3) . The perimeter of the triangle is

Correct answer: C. 2 3 + 6 & units Let the two sides a and b are the roots of the equation x^ 2 - 2 3 x + 2 = 0 . ∴ a + b = 2 3 and ab = 2 . Also, ∠ C = (π)/(3) Using cosine rule, cosC = a^ 2 + b^ 2 - c^ 2 2ab ⇒ (1)/(2) = a^ 2 + b^ 2 - c^ 2 2ab ⇒ a^ 2 + b^ 2 - c^ 2 = ab ⇒ (a + b) ^ 2 - c^ 2 = 3 ab ⇒ (2 3 ) ^ 2 - c^ 2 = 3 × 2 ⇒ c = 6 ∴ Perimeter = a + b + c = 2 3 + 6 units

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Properties of Triangles

NTA Abhyas 2020 · Mathematics · Properties of Triangles

Two sides of a triangle are given by the roots of the equation& x^...

Two sides of a triangle are given by the roots of the equation&

x^{2} - 2 \sqrt{3} \text{x} + 2 = 0

and the angle between the sides is

\dfrac{\pi}{3}

. The perimeter of the triangle is

Options

A $2 \sqrt{3}$ & units
B $\sqrt{6}$ & units
C $2 \sqrt{3} + \sqrt{6}$ & units
D $2 \left(\sqrt{3} + \sqrt{6}\right)& units$

Correct answer

C. $2 \sqrt{3} + \sqrt{6}$ & units

Step-by-step solution

Let the two sides

a

and

b

are the roots of the equation

x^{2} - 2 \sqrt{3} \text{x} + 2 = 0

∴ \text{a} + \text{b} = 2 \sqrt{3}

and

\text{ab} = 2

. Also,

∠ \text{C} = \dfrac{\pi}{3}

Using cosine rule,

\text{cosC} = \dfrac{a^{2} + b^{2} - c^{2}}{2\text{ab}} ⇒ \dfrac{1}{2} = \dfrac{a^{2} + b^{2} - c^{2}}{2\text{ab}}

⇒ a^{2} + b^{2} - c^{2} = \text{ab} ⇒ {\left(\text{a} + \text{b}\right)}^{2} - c^{2} = 3 \text{ab}

⇒ {\left(2 \sqrt{3}\right)}^{2} - c^{2} = 3 \times 2 ⇒ \text{c} = \sqrt{6}

∴

Perimeter

= \text{a} + \text{b} + \text{c} = 2\sqrt{3} + \sqrt{6}

units

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