Question

If the length of three sides of a trapezium other than base are equal to $20\text{cm}$, then find the area of trapezium when it is maximum.

Answer 1

Let $ABCD$ be the given trapezium such that $AD=DC=BC=20\text{cm}$.
$DP$ and $CQ$ are perpendiculars on $AB$.
$△APD\cong △BQC$.

Let $AP=x\text{cm}$. Then $BQ=x\text{cm}$.
By Pythagoras theorem, $DP=QC=\sqrt{400-x^2}$

Let $A$ be the area of trapezium $ABCD$. Then,
$A=\frac{1}{2}(AB+CD)\times DP=\frac{1}{2}(20+20+2x)\sqrt{400-x^2}$
$\Rightarrow A=(20+x)\sqrt{400-x^2}$
Differentiate w.r.t. $x$
$\frac{dA}{dx}=\sqrt{400-x^2}-\frac{x(20+x)}{\sqrt{400-x^2}}=\frac{400-x^2-x^2-20x}{\sqrt{400-x^2}}=\frac{400-2x^2-20x}{\sqrt{400-x^2}}$

To find the critical numbers, $\frac{dA}{dx}=0$
$\Rightarrow \frac{400-2x^2-20x}{\sqrt{400-x^2}}=0$
$\Rightarrow 400-2x^2-20x=0\Rightarrow x^2+10x-200=0$
$\Rightarrow (x-10)(x+20)=0\Rightarrow x=10,-20$
$\Rightarrow x=10$ as $x>0$

Now,
$\frac{d^{2}A}{d{x}^2}=\frac{\sqrt{400-x^2}(-20-4x)+\frac{(400-2x^2-20x)x}{\sqrt{400-x^2}}}{400-x^2}$
$=\frac{(400-x^2)(-20-4x)+(400-2x^2-20x)x}{(400-x^2{)}^{3/2}}$
$=\frac{-8000-1600x+20x^2+4x^3+400x-20x^2-2x^3}{(400-x^2{)}^{3/2}}$
$=\frac{2x^3-1200x-8000}{(400-x^2{)}^{3/2}}$

$\frac{d^{2}A}{d{x}^2}{|}_{x=10}=\frac{-18000}{{300}^{3/2}}<0$

Therefore, area of trapezium is maximum when $x=10$.
And maximum area is, $A=\frac{1}{2}(40+20)\sqrt{400-100}=300\sqrt{3}{\text{cm}}^2$