If the lengths of the three sides of a trapezium other than the base are 10 cm each, then the maximum area of the trapezium is:

$75 \sqrt{3}$ cm²

The correct answer is Option (3) → $75 \sqrt{3}$ cm²

$\text{Let the trapezium have sides }AD=10,\;BC=10,\;CD=10\text{ and base }AB=b$

$AD \parallel BC,\; CD \text{ is the upper base}$

$\text{Drop perpendiculars from }C\text{ and }D\text{ to }AB$

$\text{Let height}=h,\;\text{and projection on }AB\text{ be }x\text{ on each side}$

$\text{Then } b=10+2x$

$\text{In right triangle, }x^2+h^2=10^2$

$x^2+h^2=100$

$\text{Area }A=\frac{1}{2}(AB+CD)h=\frac{1}{2}(b+10)h$

$A=\frac{1}{2}(10+2x+10)h$

$A=(10+x)h$

$h=\sqrt{100-x^2}$

$A=(10+x)\sqrt{100-x^2}$

$\frac{dA}{dx}=\sqrt{100-x^2}+(10+x)\frac{-x}{\sqrt{100-x^2}}$

$\frac{dA}{dx}=\frac{100-x^2-(10+x)x}{\sqrt{100-x^2}}$

$\frac{dA}{dx}=\frac{100-x^2-10x-x^2}{\sqrt{100-x^2}}$

$\frac{dA}{dx}=\frac{100-2x^2-10x}{\sqrt{100-x^2}}$

$100-2x^2-10x=0$

$2x^2+10x-100=0$

$x^2+5x-50=0$

$x=\frac{-5+\sqrt{25+200}}{2}$

$x=\frac{-5+15}{2}=5$

$h=\sqrt{100-25}= \sqrt{75}=5\sqrt{3}$

$A_{\max}=(10+5)5\sqrt{3}$

$A_{\max}=75\sqrt{3}\text{ cm}^2$