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

Answer: (B)

$75\sqrt{3}\, cm^{2}$

Answer 2

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}$.