Not that angle again

One way to picture parallelogram $ABCD$ is as follows:

We can construct a segment through $C$ that is parallel to $AF$ that intersects $AE$ at $G$ , and another segment through $G$ that is parallel to $CF$ that intersects $AF$ at $H$ , as pictured below:

Then each triangle in the parallelogram is a $30°$ - $60°$ - $90°$ triangle, and using $\triangle AGH$ with segment $HG = 3$ we can determine that $AH = 2\sqrt{3}$ , and using $\triangle CEH$ with segment $CE = 6$ we can determine that $EH = 2 \sqrt{3}$ , and since $AE = AH + EH$ , $AE = 4\sqrt{3}$ . Using $\triangle ADE$ with segment $AE = 4\sqrt{3}$ we can determine that $ED = 4$ , and since $CD = CE + ED$ , $CD = 10$ . The area is therefore $10 \cdot 4\sqrt{3} = 40\sqrt{3}$ , and $40 + 3 = \boxed{43}$ .

Oh this can be done with the help of co-ordinate geometry .this will make it very simple. Anything that can be constructed stepwise with the help of protractor and scale with perfection can be solved with this method.

$Let ~AB=CD=m, ~~~~~~~AD=BC=n.\\ In ~quadrilateral ~AECF, ~opposite~ angles, \angle AEC + \angle CFA=180,\\ \therefore ~\angle FAE + \angle ECF =180 ~also.~~~ But \angle FAE=60. \\ \therefore~ \angle DAB=120,~ \angle ABE=60. ~~~So~~ \Delta~ ABE~ is ~~30-60-90,~ so~ also ~\Delta AFD.\\ \implies ~BE=\frac 1 2 m, ~~~~~FD=\frac 1 2 n.\\ \implies ~BE+EC=\color{#3D99F6}{n= \frac 1 2 m+6,~~~~similarly, ~m=\frac 1 2 n+3 }\\ Solving,~m=8,~~n=10 ~and ~height ~AE=mCos30=4*\sqrt3.\\ Area=40\sqrt3. ~~~a+b=\Large ~~~\color{#D61F06}{ \boxed{~~43~~}}$

Consider quadrilateral AECF. Angles AEC and CFA are right angles; angle EAF is 60 degrees; therefore angle ECF is 120 degrees.

It follows that angles B and D of the parallelogram are 60 degrees each.

Let BC = x and CD = y.

Consider triangle ABE, which is a 30-60-90 triangle. We have AB = 2 BE = 2 (x - 6). On the other hand, AB = CD = y. Likewise, from triangle ADF we have AD = 2 DF = 2 (y - 3), and AD = CB = x.

Thus we have the equations 2 (x-6) = y and 2 (y-3) = x. Solving gives x = 10 and y = 8.

The heights follow from triangles ABE and ADF: h = BE = (x-6) sqrt 3 = 4 sqrt 3, and k = DF = (y-3) sqrt 3 = 5 sqrt 3.

The area is A = x h = y k = 40 sqrt 3. The solution is therefore 40 + 3 = 43.

Let x be the length of BC. Let y be the length of CD. Since CE = 6  BE = x – 6. Since CF = 3  DF = y - 3. In the figure AECF we have: Angle A = 60, angle F = 90, angle E = 90  angle BCD = 120  angle B = angle D = 60. In triangle ABE, we have: BE/AB = cos 60 = ½  (x – 6)/y = ½  y = 2x – 12 (1) In triangle AFD, we have: DF/AD = cos 60 = ½  (y – 3)/x = ½  x = 2y – 6 (2) Solving equations (1) and (2) gives: x = 10 and y = 8. In triangle ABE, we have: AE/AB = sin 60 = √3/2  AE = 4√3 The area of the parallelogram is then: AE X BC = 10 X 4√3 = 40√3 That means a = 40 and b = 3. Therefore a + b = 40 + 3 = 43.