Question

Rectangle of maximum area that can be inscribed in an equilateral triangle of side $a$ will have area =

• A. $\frac{a^2\sqrt{3}}{2}$
• B. $\frac{a^2\sqrt{3}}{4}$
• C. $\frac{a^2\sqrt{3}}{8}$
• D. none

Answer 1

The correct option is C $\frac{a^2\sqrt{3}}{8}$

Let the side $BC=a$ be chosen along x-axis and altitude AD be along y-axis.

$A{D}^2=A{C}^2-D{C}^2=a^2-\frac{a^2}{4}=\frac{3a^2}{4}$ $\therefore AD=\frac{a\sqrt{3}}{2}$

Let QPSR be the rectangle inscribed in the triangle.

If $A$ be its area, then $A=2xy$ where $(x,y)$ are the co-ordinates of vertex P which lies on line AC whose equation in intercept form is $\frac{x}{a/2}+\frac{y}{\sqrt{3a/2}}=1$ or $\frac{2x}{a}+\frac{2y}{a\sqrt{3}}=1\cdots \left(1\right)$

Area $A=2xy=x(1-\frac{2x}{a})a\sqrt{3}$ ...[ from (1) ]

$\frac{dA}{dx}=a\sqrt{3}(1-\frac{4x}{a})=0$

$\therefore x\frac{a}{4}$ $\frac{d^{2}A}{d{x}^2}=-$ ive and A is maximum.

$\therefore A=\frac{a}{4}(1-\frac{1}{2})a\sqrt{3}=\frac{a^2\sqrt{3}}{8}$

Ans: C