Question 1210155
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{{{highlight(cross(ABCE))}}} <U>ABCF</U> is a Parallelogram. ADEF is a Rhombus. Find the area of the composite figure.
https://ibb.co/Jj2jKPD3
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;I edited your post to turn  your incorrect writing into a correct problem.



<pre>
In rhombus ADEF, the half of the long diagonal is  {{{sqrt(17^2-5^2)}}} = {{{sqrt(264)}}} = {{{2*sqrt(66)}}} units long.


So, we can find the area of the parallelogram ABCF as  the product of the base FC = 11+5+5 = 21 by the height AG = {{{2*sqrt(66)}}}.


So, the area of the parallelogram ABCF is  {{{21*(2*sqrt(66))}}} = {{{42*sqrt(66)}}}.


To it, we should add the area of triangle FDE  {{{(1/2)*(5+5)*GE}}} = {{{5*2*sqrt(66)}}} = {{{10*sqrt(66)}}}.


Thus, the total combined area is {{{42*sqrt(66)}}} + {{{10*sqrt(66)}}} = {{{52*sqrt(66)}}}.


<U>ANSWER</U>.  The total combined area is  {{{52*sqrt(66)}}}  square units.
</pre>

Solved.


As simple as a cucumber.



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Edwin, you do not need to revise your solution - it is correct.


The solution by @CPhill is wrong.