Question

The three sides of a trapezium are equal, each being $8cm$. The area of the trapezium, when it is maximum is ?

• A. $24\sqrt{3}c{m}^2$
• B. $48\sqrt{3}c{m}^2$
• C. $72\sqrt{3}c{m}^2$
• D. None of the above.

Answer 1

The correct option is B $48\sqrt{3}c{m}^2$

Area of Trapezium $=AB\times AE+2\times \frac{1}{2}(DE\times AE)$

$△=8\times 8\sin \theta +\left(8\cos \theta \times 8\sin \theta \right)$

$=64\sin \theta +64\cos \theta \sin \theta$

$=64\sin \theta +64\frac{\sin 2\theta }{2}$

$△=64\sin \theta +32\sin 2\theta$

Differentiate w.r.t $\theta$

$\frac{d△}{d\theta }=64\cos \theta +2\times 32\cos 2\theta$

Equate $\frac{d△}{d\theta }=0$

$64\cos \theta +2\times 64\cos 2\theta =0$

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

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

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

$(2\cos \theta -1)(\cos \theta +1)=0$

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

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

Hence 0 cannot be $\pi$ or else the Triangle ADC would not been formed.

Since $\theta$ is an angle of Triangle $\theta <\pi$

$\therefore \theta =\frac{\pi }{3}$

Therefor area is maximum when $\theta =\frac{\pi }{3}$

$△=64\sin \theta +32\sin 2\theta$

$=64\sin \left(\frac{\pi }{3}\right)+32\sin \left(\frac{\pi }{3}\right)$

$=64\frac{\sqrt{3}}{2}+32\frac{\sqrt{3}}{2}$

$△=48\sqrt{3}{cm}^2$

Hence option B is correct.