Question

Let $A(1,-1),B(4,-2)$ and $C(9,3)$ be the vertices of the triangle $ABC$. A parallelogram AFDE is drawn with vertices $D,E$ and $F$ on the line segments $BC,CA$ and $AB$ respectively. Find the maximum area of parallelogram $AFDE$.

Answer 1

Answer: 5

Let $AF=x=DE$ and $AE=y=DF$
As $△CAB$ is $△CED$
So, $\frac{CE}{CA}=\frac{DE}{AB}\Rightarrow \frac{b-y}{b}=\frac{x}{c}\Rightarrow y=b\left(1-\frac{x}{c}\right)$
(Here $BC=a,AC=b$ and $AB=c$ )
Now, area of parallelogram AFDE
$=S=(AF)(EM)=xy\sin A\Rightarrow S=x\cdot b\left(1-\frac{x}{c}\right)\sin A$ (Note $:\sin A$ is fixed).....(1)
Now, differentiating both sides of equation (1) with respect to $x$, we get
$\frac{dS}{dx}=\frac{b}{c}(c-2x)\sin A=0\Rightarrow x=\frac{c}{2}$
Also, ${\frac{d^{2}s}{d{x}^2}]}_{x=\frac{c}{2}}=\frac{-2b}{c}<0$
So, $S$ is maximum when $x=\frac{c}{2}$
Now, $S_{\max }=\frac{1}{4}bc\sin A$

$=\frac{1}{2}\left(\frac{1}{2}bc\sin A\right)=\frac{1}{2}(\mathrm{area}(△ABC))=\frac{1}{4}\left|\begin{array}{ccc}1& -1& 1\\ 4& -2& 1\\ 9& 3& 1\end{array}\right|\mid =\frac{20}{4}=5$ (square units.)