Single Choice

A rectangle is inscribed in an equilateral triangle of side length $$2a$$ units. The maximum area of this rectangle can be

A$$\sqrt 3a^2$$
B$$\dfrac {\sqrt 3a^2}4$$
C$$\dfrac {\sqrt 3a^2}2$$
Correct Answer
D$$a^2$$

Solution

In $$\triangle DBF,\tan { 60° } =\cfrac { 2b }{ 2a-l } $$
$$\sqrt { 3 } =\cfrac { 2b }{ 2a-l } $$
$$b=\cfrac { \sqrt { 3 } \left( 2a-l \right) }{ 2 } $$
A, Area of rectangle $$DEFG=l\times b$$
$$=l\times \cfrac { \sqrt { 3 } }{ 2 } (2a-l)$$
For maximum area
$$\cfrac { dA }{ dl } =0$$
$$\cfrac { \sqrt { 3 } }{ 2 } \left( 2a-2l \right) =0$$
$$a=l$$
Therefore, Maximum Area$$=l\times b$$
$$=a\times \cfrac { \sqrt { 3 } }{ 2 } (2a-a)$$
Maximum Area$$=\cfrac { \sqrt { 3 } { a }^{ 2 } }{ 2 } $$ (Answer)


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