Inscribed Trapezoid

As shown above, label the center of the semicircle as $O$ , draw radius $OC$ (length $2)$ , draw the altitude to $AD$ from point $C$ to $E$ , label $CD$ as $1$ , $OE$ as $x$ , and $ED$ as $2-x$ . Begin by observing that altitude $CE$ constructs right triangles $OEC$ and $DEC$ . Using the Pythagorean Theorem , we can solve for $x$ as such:

$CE^{2}=2^{2}-x^{2}$

$CE^{2}=1^{2}-(2-x)^{2}$

$\therefore 1^{2}-(2-x)^{2}$ = $2^{2}-x^{2}$ $\Rightarrow$ $1-(2-x^{2}=4-x^{2}$ $\Rightarrow$ $1-(2-x)(2-x)=4-x^{2}$ $\Rightarrow$ $1-(4-4x+x^{2})=4-x^{2}$ $\Rightarrow$ $1+(-4+4x-x^{2})=4-x^{2}$ $\Rightarrow$ $-3+4x-x^{2}+x^{2}=4$ $\Rightarrow$ $4x=7$ $\Rightarrow$ $\boxed{x=\frac{7}{4}}$

Now that we have solved for $x$ , we can substitute $\frac{7}{4}$ for $x$ to solve for altitude $CE$ as such:

$CE^{2}=2^{2}-(\frac{7}{4})^{2}$ $\Rightarrow$ $CE^{2}=4-\frac{49}{16}$ $\Rightarrow$ $CE^{2}=\frac{15}{16}$ $\Rightarrow$ $\boxed{CE=\frac{\sqrt{15}}{4}}$

Now that we have solved for the altitude/height $(CE)$ of Trapezoid ABCD , we can solve for the area of Trapezoid ABCD as such:

Area of Trapezoid $\Rightarrow$ $\frac{b_{1}+b_{2}}{2} \times h$

$\therefore A_{Trapezoid ABCD}=\frac{(AD)+(BC)}{2} \times {(CE)}$ $\Rightarrow$ $A_{Trapezoid ABCD}=\frac{(4+7\sqrt{2})}{2} \times {\frac{\sqrt{15}}{4}}$ $\Rightarrow$ $A_{Trapezoid ABCD}=\frac{15\sqrt{15}}{16}$

$\frac{15\sqrt{15}}{16}\approx\boxed{{3.631}}$

Connect $BD$ , $\angle ABD=90^{\circ}$ ,so $BD=\sqrt{15}$ .

Let $E$ be on $AD$ and $\angle AEB=90^{\circ}$ ,then $\triangle ABE\text{~}\triangle ADB$ .

So, $AE=\frac14,BC=4-\frac14\times2=3.5$ ,and $BE=\frac{\sqrt{15}}{4}$

Hence the area is $(4+3.5)\times\frac{\sqrt{15}}{4}\times\frac12=\frac{15\sqrt{15}}{16}$

Apply pythagorean theorem on $\triangle ABE$ . We get

$h^2=1-x^2$ $(1)$

Apply pythagorean theorem on $\triangle BEF$ . We get

$h^2+x^2-4x=0$ $(2)$

Substituting $(1)$ in $(2)$ then simplifying, we get

$x=\dfrac{1}{4}$

It follows that

$h=\dfrac{\sqrt{15}}{4}$ and $b=\dfrac{7}{2}$

The area of a trapezoid is $\dfrac{1}{2}(base~1~+base~2) \times height$ . Substitute:

$A=\dfrac{1}{2}\left(\dfrac{7}{2}+4\right)\left(\dfrac{\sqrt{15}}{4}\right) \approx \boxed{3.631}$