Question

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. $\frac{\sqrt{3}{a}^2}{4}$
• C. $\frac{\sqrt{3}{a}^2}{2}$
• D. $a^2$

Answer 1

The correct option is C $\frac{\sqrt{3}{a}^2}{2}$

In $△DBF,\tan 60°=\frac{2b}{2a-l}$

$\sqrt{3}=\frac{2b}{2a-l}$

$b=\frac{\sqrt{3}(2a-l)}{2}$

A, Area of rectangle $DEFG=l\times b$

$=l\times \frac{\sqrt{3}}{2}(2a-l)$

For maximum area

$\frac{dA}{dl}=0$

$\frac{\sqrt{3}}{2}(2a-2l)=0$

$a=l$

Therefore, Maximum Area$=l\times b$

$=a\times \frac{\sqrt{3}}{2}(2a-a)$

Maximum Area$=\frac{\sqrt{3}{a}^2}{2}$ (Answer)