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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`

A

`sqrt(3a^(2))`

B

`sqrt(3a^(2))/(4)`

C

`a^(2)`

D

`sqrt(3a^(2))/(2)`

Text Solution

Verified by Experts

The correct Answer is:
4

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)
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