Question

The three sides of a trapezium are equal, each being 6" long. Find the area of the trapezium when it is maximum.

• A. $18\sqrt{3}$ sq. inch.
• B. $54$ sq. inch.
• C. $27\sqrt{3}$ sq. inch.
• D. $71$ sq. inch.

Answer 1

Answer: (C)

$27\sqrt{3}$ sq. inch.

$AP=BQ=6\sin \alpha$

$DP=QC=6\cos \alpha .$

$A=$ area of the trapezium $\frac{1}{2}[AB+DC].BQ$

$=\frac{1}{2}(6+6+12\cos \alpha ).6\sin \alpha$

$=36\sin \alpha (1+\cos \alpha )$

A will be maximum when $\alpha =\pi /3$

$\therefore A=36(\sin {60}^{\circ }+\frac{1}{2}\sin {120}^{\circ })$

$=36[\frac{\sqrt{3}}{2}+\frac{1}{2}.\frac{\sqrt{3}}{2}]=27\sqrt{3}$ sq. inch.
Ans: C