The maximum area of a parallelogram inscribed | vedclass

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(C) Let $AF = x$. Since $FM \parallel ED$ and $FE \parallel AD$,the parallelogram is $AMFE$ (or $AFED$ based on the diagram). Let $AF = x$. Since $\De

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$\frac{bc}{4} \sin A$

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(C) Let $AF = x$. Since $FM \parallel ED$ and $FE \parallel AD$,the parallelogram is $AMFE$ (or $AFED$ based on the diagram). Let $AF = x$. Since $\Delta AFE \sim \Delta ABC$,we have $\frac{FE}{BC} = \frac{AF}{AB} = \frac{AE}{AC}$.Let $AF = x$. Then $FE = \frac{x}{b} \cdot c$. The height of the parallelogram with respect to base $AF$ is $h = x \sin A$.The area of the parallelogram is $Area = 	ext{base} 	imes 	ext{height} = FE 	imes (x \sin A) = (\frac{cx}{b}) 	imes (x \sin A) = \frac{c \sin A}{b} x^2$. However,looking at the diagram,the base is $AD$ and height is $FM$. Let $AF = x$. Then $FE = \frac{x}{b} c$. The height of the parallelogram from $FE$ to $AB$ is $x \sin A$. Area $A(x) = FE \cdot (x \sin A) = (\frac{c}{b} x) \cdot (x \sin A) = \frac{c \sin A}{b} x^2$. This is not correct as $x$ is limited. Correct approach: Let $AF = x$. Then $FE = \frac{c}{b}(b-x)$. The height of the parallelogram is $h = x \sin A$.Area $A(x) = FE \cdot h = \frac{c}{b}(b-x) \cdot x \sin A = \frac{c \sin A}{b} (bx - x^2)$.To maximize,$\frac{dA}{dx} = \frac{c \sin A}{b} (b - 2x) = 0 \Rightarrow x = \frac{b}{2}$.Maximum Area $= \frac{c \sin A}{b} (b(\frac{b}{2}) - (\frac{b}{2})^2) = \frac{c \sin A}{b} (\frac{b^2}{4}) = \frac{bc}{4} \sin A$.

The maximum area of a parallelogram inscribed in a triangle $ABC$ (given that $A$ is also one of the vertices of the parallelogram) is equal to (where symbols have their usual meaning in $\Delta ABC$):

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