Question

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

• A. 27sq. inch
• B. 21sq. inch
• C. $27 \sqrt { 3 }$sq.inch
• D. ) None of these

Answer 1

Answer: (C)

$27 \sqrt { 3 }$sq.inch

Answer 2

Area of trapezium ABCD

$S = \frac { 1 } { 2 } [ 6 + ( 6 + 2 \times 6 \cos \theta ) ] \times 6 \sin \theta$

= 36$( \sin \theta + \sin \theta \cos \theta )$

= $36 \sin \theta + 18 \sin 2 \theta$

$\frac { d s } { d \theta } = 36 \cos \theta + 36 \cos 2 \theta = 0$

= $36 ( \cos \theta + \cos 2 \theta ) = 0$

⇒ $\cos \theta + 2 \cos ^ { 2 } \theta - 1 = 0$

$2 \cos ^ { 2 } \theta + \cos \theta - 1 = 0$

$\cos \theta = - 1$or $\cos \theta = \frac { 1 } { 2 }$

$\theta = \pi$ or $\frac { \pi } { 3 }$ $\theta \neq \pi$

$S _ { \max }$ at $\theta = \frac { \pi } { 3 }$

$36 \sin \frac { \pi } { 3 } + 18 \sin \frac { 2 \pi } { 3 }$

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

So 'C' is correct