Question

Let the three sides of a trapezium are equal and each equal to $6\,cm$ . If area of trapezium is maximum, then length of fourth side of trapezium is

• A. $6\,cm$
• B. $9\,cm$
• C. $12\,cm$
• D. $15\,cm$

Answer 1

Answer: (C)

$Ar=\frac{1}{2}\left(\right.12+2x\left.\right)\cdot \sqrt{36 - x^{2}}$
$\frac{d A}{d x}=1\cdot \sqrt{36 - x^{2}}+\frac{\left(\right. 6 + x \left.\right) \left(\right. - 2 x \left.\right)}{2 \sqrt{36 - x^{2}}}=0$

$\Rightarrow 36-x^{2}-x^{2}-6x=0$
$\Rightarrow x^{2}+3x-18=0$
$\Rightarrow \left(\right.x+6\left.\right)\left(\right.x-3\left.\right)=0$
$\Rightarrow x=3$