Question

If the length of $3$ sides of a trapezium other than base are equal to $10$ cm , then find the area of the trapezium when it is maximum.

Answer 1

Consider the trapezium $ABCD$, where $AD\parallel BC$, $AB=CD=AD=10am$

Let $BC$ be $x$

Draw $AM\perp BC$ and $DN\perp BC$

$\therefore$ $MN=10cm$

Let $BC=xcm$

Now, $CN=\left(\frac{x-10}{2}\right)cm=(\frac{x}{2}-5)cm$

Using Pythagoras theorem in $△DNC$,

$DN=\sqrt{C{D}^2-C{N}^2}=\sqrt{(10{)}^2-{(\frac{x}{2}-5)}^2}cm$

Thus, Area of trapezium $ABCD=\frac{1}{2}(BC+AD)\times DN=\frac{1}{2}(x+10)\sqrt{(10{)}^2-{(\frac{x}{2}-5)}^2}c{m}^2$

Let $A=\frac{1}{2}(x+10)\sqrt{(10{)}^2-{(\frac{x}{2}-5)}^2}c{m}^2$

$\Rightarrow$ $A^2=\frac{1}{4}(x+10{)}^2\left[\right(10{)}^2-{(\frac{x}{2}-5)}^2]c{m}^2$

$\Rightarrow$ $A^2=\frac{1}{4}(x+10{)}^2[10+(\frac{x}{2}-5)][10-(\frac{x}{2}-5)]$

$\Rightarrow$ $A^{=}\frac{1}{4}(x+10{)}^2(\frac{x}{2}+5)(15-\frac{x}{2})$

$\Rightarrow$ $A^2=\frac{1}{16}(x+10{)}^3(30-x)$

$\Rightarrow$ Let $P=16A^2=(x+10{)}^3(30-x)$

More is $P$, more is $A$

Thus, area is more

$\frac{dP}{dx}=(x+10{)}^3(-1)+3(x+10{)}^2(30-x)$

$=(x+10{)}^2[-(x+10)+3(30-x)]$

$=(x+10{)}^2(80-4x)$

$=4(x+10{)}^2(20-x)$

$\frac{d^{2}P}{d{x}^2}=4\left[2\right(x+10\left)\right(20-x)+(x+10{)}^2(-1)]$

$=4(x+10)\left[2\right(20-x)-(x-10\left)\right]$

$=4(x+10)(30-3x)$

$=12(x+10)(10-x)$

$\frac{dP}{dx}=0$ Implies that $x+10=0$ or $20-x=0$

$\Rightarrow$ $x=-10$ or $x=20$

$x$ cannot be negative

$\therefore$ $x=210$

Now, for $x=20,\frac{d^{2}P}{d{x}^2}=12\times 30\times (-10)=-3600<0$

Thus, $P$ is maximum for $x=20$

Thus, maximum area of the trapezium $=\frac{1}{4}\sqrt{(20+10{)}^3(30-20)}$

$=\frac{1}{4}\times \sqrt{(30{)}^3\times 10}c{m}^2$

$=\frac{1}{4}\times 30\times 10\sqrt{3}c{m}^2=75\sqrt{3}c{m}^2$