Question ID: #1039
Let $A(4,-2)$, $B(1,1)$ and $C(9,-3)$ be the vertices of a triangle $ABC$. Then the maximum area of the parallelogram $AFDE$, formed with vertices $D$, $E$ and $F$ on the sides $BC$, $CA$ and $AB$ of the triangle $ABC$ respectively, is
Solution:
Area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is:
$$ \Delta = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right| $$
$$ \text{Area of } \Delta ABC = \frac{1}{2} \left| \begin{vmatrix} 4 & -2 & 1 \\ 1 & 1 & 1 \\ 9 & -3 & 1 \end{vmatrix} \right| $$
$$ \Delta ABC = \frac{1}{2} | 4(1 – (-3)) – (-2)(1 – 9) + 1(-3 – 9) | $$
$$ \Delta ABC = \frac{1}{2} | 4(4) + 2(-8) + 1(-12) | $$
$$ \Delta ABC = \frac{1}{2} | 16 – 16 – 12 | $$
$$ \Delta ABC = \frac{1}{2} |-12| = 6 \text{ sq. units} $$
The maximum area of a parallelogram $AFDE$ inscribed in $\Delta ABC$ (sharing angle $A$ with the triangle) is exactly half the area of $\Delta ABC$.
$$ \text{Area of } AFDE = \frac{1}{2} \times \text{Area of } \Delta ABC $$
$$ \text{Area of } AFDE = \frac{1}{2} \times 6 = 3 \text{ sq. units} $$
Ans. (3)
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