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If the length of each side of an equilateral triangle is increased by 2 units, the area is found to be increased by 3 + √3 square unit. The length of each side of the triangle is
Answer & Solution
Correct Answer:
Option
A
Let each side of the triangle be a units
$$\eqalign{ & \Rightarrow \frac{{\sqrt 3 }}{4}\left\{ {{{\left( {a + 2} \right)}^2} - {a^2}} \right\} = 3 + \sqrt 3 \cr & \frac{1}{4}\left( {{a^2} + 4 + 4a - {a^2}} \right) = 1 + \sqrt 3 \cr & \frac{1}{2}\left( {4 + 4a} \right) = 1 + \sqrt 3 \cr & 1 + a = 1 + \sqrt 3 \cr & a = \sqrt 3 {\text{ units}} \cr} $$
$$\eqalign{ & \Rightarrow \frac{{\sqrt 3 }}{4}\left\{ {{{\left( {a + 2} \right)}^2} - {a^2}} \right\} = 3 + \sqrt 3 \cr & \frac{1}{4}\left( {{a^2} + 4 + 4a - {a^2}} \right) = 1 + \sqrt 3 \cr & \frac{1}{2}\left( {4 + 4a} \right) = 1 + \sqrt 3 \cr & 1 + a = 1 + \sqrt 3 \cr & a = \sqrt 3 {\text{ units}} \cr} $$
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