Question

If the length of three sides of a trapezium other than base are equal to $10 \,cm$, then find the area of trapezium when it is maximum.

• A. $75\sqrt{3}/2\, cm^{2}$
• B. $75\sqrt{3}\, cm^{2}$
• C. $90\sqrt{3}\, cm^{2}$
• D. $48\, cm^2$

Answer 1

The Correct Option is B

Let $ABCD$ be the given trapezium such that $AD = DC = BC = 10 \,cm$. Draw $DP$ and $CQ$ perpendiculars from $D$ and $C$ respectively on $AB$. Clearly, $\Delta APD ? Let $AP = x \,cm$. Then, $BQ = x\, cm$. By applying Pythagoras Theorem in $\Delta APD$ and $\Delta BQC$, we have $DP = QC = \sqrt{100-x^{2}}$ Let $A$ be the area of trapezium $ABCD$. Then, $A = \frac{1}{2}\left(10+10+2x\right)\times \sqrt{100-x^{2}}$ $\Rightarrow A = \left(10+x\right)\sqrt{100-x^{2}}$ $\Rightarrow \frac{dA}{dx} = \sqrt{100-x^{2}} - \frac{x\left(10+x\right)}{\sqrt{100-x^{2}}}$ $= \frac{100-10x-2x^{2}}{\sqrt{100-x^{2}}}$ For maximum or minimum values of $A$, we must have $\frac{dA}{dx} = 0$

$\Rightarrow \frac{100-10x-2x^{2}}{\sqrt{100-x^{2}}} = 0$ $\Rightarrow 100 - 10x-2x^{2} = 0$ $\Rightarrow x^{2} + 5x - 50 = 0$ $\Rightarrow \left(x + 10\right)\left(x - 5\right) = 0$ $\Rightarrow x = 5 \left[\because x > 0 \therefore x + 10 \ne 0\right]$ Now, $\frac{d^{2}A}{dx^{2}} = \frac{\sqrt{100-x^{2}}\left(-10-4x\right)+\frac{\left(100-10x-2x^{2}\right)x}{\sqrt{100-x^{2}}}}{100-x^{2}}$ $\Rightarrow \frac{d^{2}A}{dx^{2}} = \frac{2x^{3} - 300x -1000}{\left(100-x^{2}\right)^{3/2}}$ $\Rightarrow \left(\frac{d^{2}A}{dx^{2}}\right)_{x = 5} < 0$ Thus, the area of the trapezium is maximum when $x = 5$. The maximum area is given by $A = \left(10 + 5\right)\sqrt{100-25}$ $=75\sqrt{3}\,cm^{2}$.