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A rectangle of the greatest area is inscribed in a trapezium ABCD, one of whose non-parallel sides AB is perpendicular to the base, so that one of the rectangle’s side lies on the larger base of the trapezium. The base of trapezium are 6 cm and 10 cm and AB is 8 cm long. Then the maximum area of the rectangle is.

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$36 c m^{2}$

$70 c m^{2}$

$24 c m^{2}$

$48 c m^{2}$

answer is B.

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Detailed Solution

Let rectangle BEFG is inscribed its area, $A = x y$

Now $Δ F E C$ and $Δ D H C$ are similar i.e.,

$\begin{array}{l} \frac{x}{8} = \frac{10 - y}{4} \Rightarrow y = 10 - \frac{x}{2} \\ \Rightarrow A = x (10 - \frac{x}{2}) w h e r e x \in (0, 8] \end{array}$

Now, $\frac{d A}{d x} = 10 - x,$ For $x \in (0, 8), \frac{d A}{d x} > 0,$ i.e., A increases

Hence, $A_{\max}$ occurs when $x = 8 .$ Hence, max area

$A_{\max} = 8 (10 - \frac{8}{2}) = 48 c m^{2}$

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