Tangent Trapezoid

Here is the solution.

A and B are the centers of the circles. Tangent chord ED cut line of centers AB at 90 degrees at H. Tangent chord FC cut line of centers AB at 90 degrees at K.

$Draw~ BG = ~and~ || EF. ~~ So~~ FG=EB~ and~ 90~ degree~ at~ G\\and~ ~AG=12-3=9. \\$
$rt. ~\angle d~ \Delta AGB,~ AB=3+12=15, ~AG=9~\therefore~BG=12. \\\angle~BAG=\phi. ~~~~~~Sin\phi=\dfrac 4 5 , ~~~~~~Cos\phi=\dfrac 3 5 . \\ rt. ~\angle d~ \Delta AFH,~ AH=AFCos\phi=12*\dfrac 3 5, ~~FH=AFSin\phi=12*\dfrac 4 5\\ rt. ~\angle d~ \Delta BEK,~ BK=EBCos\phi=3*\dfrac 3 5, ~~EK=BESin\phi=3*\dfrac 4 5\\Area~trapezium~FEKH=\dfrac 1 2 (EK+FH)*(HK)\\ = \dfrac 1 2 (EK+FH)*(AB-AH+BK)\\= \dfrac 1 2 (3*\dfrac 4 5+12*\dfrac 4 5)(15-12*\dfrac 3 5+3*\dfrac 3 5)= 57.6.\\\therefore ~area~~~ CDEF =2*57.6 =~~~~ \color{#D61F06}{115.2}$

Focus on $OQEF$ that is rectangle trapezoid and $OQ= 15 \rightarrow 15^2=(12-3)^2+ h^2 \rightarrow h= 12 \therefore FE=CD=12$

Let $m=DX$ , we have that $\triangle OCX \sim \triangle QDX$ and $OC||DQ$ so $\frac{12+m}{m}=\frac{12}{3} \rightarrow m=4$

Let $h_1$ height of $\triangle QDX$ since $D \rightarrow DE=2h_1$ so $\frac{h_1 \times 5}{2}=\frac{3\times 4}{2} \rightarrow h_1=\frac{12}{5}$

Let $h_2$ height of $\triangle OCX$ since $C \rightarrow CF=2h_2$ so $\frac{h_2 \times 20}{2}=\frac{12\times 16}{2} \rightarrow h_2=\frac{48}{5}$

$DE=\frac{24}{5}$

$CF=\frac{96}{5}$

As trapezoid $CDEF$ is isosceles and let $h_3$ its height. $CD^2=(\frac{CF-DE }{2})^2+{h_3}^2 \rightarrow h_3=\sqrt{12^2-(7.2 )^2}=9.6$

Area of the trapezoid $CDEF=\frac{(DE+CF)h_3}{2}=\frac{(\frac{24}{5}+\frac{96}{5})(9.6)}{2}=\boxed{115.2}$

First i will find the generalization the problem.

Let the radius of the circles be $a$ and $b$ .Let $[KLM...XYZ]$ denote the area of polygon $KLM...XYZ$ .

Let the common internal tangent intersect $CD$ and $EF$ at points $P$ and $R$ respectively. Let $Q$ be the intersection of $AB$ and $PR$ . Tangents to a circle from an external point have the same length. Therefore,

$PC = PQ = PD$

and

$RE = RQ = RF$

By symmetry, $PQ = RQ$ . Therefore,

$CD = PR = EF$

Construct $PS$ perpendicular to $EF$ with $S$ on $EF$ and $AG$ perpendicular to $BE$ with $G$ on $BE$ .

Triangles $PSR$ and $AGB$ are similar since corresponding sides are perpendicular to each other. Therefore,

$\dfrac{PS}{PR} = \dfrac{AG}{AB}$

Since $AG = EF$ , we have

$EF^2 = AB^2 - BG^2 = (b + a)^2 - (b - a)^2 = 4ab$

and

$PS = \dfrac{EF^2}{AB} = \dfrac{4ab}{(a + b)}$

See first

$\dfrac{CP}{PD} = \dfrac{ER}{RF}$ implies

$[CDEF] = [PEF] + [RCD] = 2[PEF]$

$= PS.EF = \dfrac{8ab\sqrt{ab}}{(a + b)}$ .

So put $a = 3$ and $b = 12$ in the above expression to get the Area $A = 115.2$

(1/ 2)(a + b)

= (1/ 2) [2 R Sin q + 2 r Sin q]

= (R + r) [2 Sqrt (R r)/ (R + r)] = 2 Sqrt (R r)

h

= r (R - r)/ (R + r) + R + r - R (R - r)/ (R + r)

= 4 R r/ (R + r)

Area of trapezium

= (1/ 2)(a + b) h

= 8 R r Sqrt (R r)/ (R + r)

= 8 (12)(3) Sqrt (12 x 3)/ (12 + 3) = 1728/ 15 = 115.2

Note: Cos q = (R - r)/ (R + r) and hence Sin q = 2 Sqrt (R r)/ (R + r) while both circles applied the same angle in calculations.