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Cartesian plane

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Question

The maximum area of the convex polygon formed by joining the points $A (0, 0), B (2 t^{2}, 0), C (18, 2), D (\frac{8}{r^{2}}, 4)$ and $E (0, 2),$ where $t \in R - \{0\}$ and interior angle at vertex B is greater than or equal to $90^{0}$ is

Moderate

Solution

Slope of $BC = \frac{2 - 0}{18 - 2 t^{2}} \geq 0$
$\Rightarrow 18 - 2 t^{2} \geq 0$
$\Rightarrow t^{2} - 9 \leq 0 \Rightarrow t \in [- 3, 3]$
Now area of quadrilateral
$=$ Area of trapezium $ABCE + Area of triangle CDE$
$= \frac{1}{2} (2 t^{2} + 18) \times 2 + \frac{1}{2} \times 18 \times 2$
$= 36 + 2 t^{2} \leq 54$
So, the required maximum area is 54 aq. units

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