A rectangle is inscribed in an equilateral triangle, with one side on a side of the triangle. If the triangle has side of length 2, what is the maximum possible area of the rectangle?

1 Answer
CW
May 25, 2018

#max. A= sqrt3/2 " units"^2#

Explanation:

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let #b and h# be the base and the height of the rectangle #DEFG#, respectively, as shown in the figure.
Given that the equilateral triangle #DeltaABC# has side of length #2, => b+2x=2#,
# => x=(2-b)/2=1-b/2#
#h/x=tan60, => h=xtan60=sqrt3x#
Area of rectangle #DEFG=|DEFG|=b*h#
#=> |DEFG|=bsqrt3x=sqrt3b(1-b/2)#
#=sqrt3(b-b^2/2)#
#= sqrt3/2(2b-b^2)#
#= -sqrt3/2(b^2-2b)#
#=-sqrt3/2((b-1)^2-1)#
#=sqrt3/2-sqrt3/2(b-1)^2#
#=> max|DEFG|# can be obtained when #b=1#
#=> max. |DEFG|=sqrt3/2 " units"^2#

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