Question

Three sides of a trapezium are equal, each being a cms long. Then its maximum area is

• A. $\frac{3\sqrt{3}}{4}{a}^2$
• B. $\frac{6\sqrt{3}}{4}{a}^2$
• C. $\frac{\sqrt{3}}{4}{a}^2$
• D. $\frac{\sqrt{3}}{2}{a}^2$

Answer 1

Answer: (A)

$\frac{3\sqrt{3}}{4}{a}^2$

$area=x\sqrt{a^2-x^2}+a\sqrt{a^2-x^2}$
$\frac{d\left(area\right)}{dx}=(x+a)\left(\sqrt{a^2-x^2}\right)$
$0=\sqrt{a^2-x^2}-\frac{x(x+a)}{\sqrt{a^2-x^2}}$
$0=(a^2-x^2)-(x^2+ax)$
$=a^2-ax-2x^2$
$0=2x^2+ax-a^2$
$=2x^3+2ax-ax-a^2$
$=2x(x+a)-a(x+a)$
$x=-a$ $x=\frac{a}{2}$
$area=(\frac{a}{2}+a)\left(\sqrt{a^2-\frac{a^2}{4}}\right)$
$=\frac{3}{2}a\frac{\sqrt{3}}{2}a$