Question

If length of three sides of a trapezium other than base are equal to $5cm$, then the base length of the trapezium when the area is maximum is

• A. $5$
• B. $10$
• C. $2.5$
• D. $7.5$

Answer 1

The correct option is B $10$


Area of the trapezium
$A=\frac{1}{2}\times (5+5+2x)\times \left(\sqrt{25-x^2}\right)\Rightarrow A=(5+x)\times \left(\sqrt{25-x^2}\right)\Rightarrow \frac{dA}{dx}=\left(\sqrt{25-x^2}\right)+\frac{(5+x)(-2x)}{2\sqrt{25-x^2}}$
Maximizing the area $\frac{dA}{dx}=0$
$2(25-x^2)-10x-2x^2=0\Rightarrow -4x^2-10x-50=0\Rightarrow 2x^2+10x-25=0\Rightarrow (2x-5)(x+5)=0\Rightarrow x=\frac{5}{2}$
Hence the length of the base is $10cm$