A rectangle is inscribed in an equilateral triangle of side length `2a` units. The maximum area of this rectangle can be (a) `sqrt(3)a^2` (b) `(sqrt(3)a^2)/4` `a^2` (d) `(sqrt(3)a^2)/2`

Options

A. `sqrt(3a^(2))`
B. `sqrt(3a^(2))/(4)`
C. `a^(2)`
D. `sqrt(3a^(2))/(2)`

Answer

4

Explanation

Let `BD_(1)=x or BC_(1)=(a-x)`
`therefore BC = (a-x)tan(pi)/(3)=sqrt(a-x)`
Now area of rectangle ABCD
`triangle =(AB)(BC)=2 sqrt(3x(a-x))`
`lt 2 sqrt(3)(x+a-x)/(2)^(2)=sqrt (3a^(2))/(2)` (using A.M ge GM)