Question 1061892
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In the given figure, the two rectangles EFGH and ACDE share a common corner at E and overlap so that BC = 7. 
What is the area of the shaded region ABGFEA?
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<pre>
1.  Consider the quadrilateral ABHE (which is the intersection area of the original rectangles).


2.  Draw the diagonal BE in this quadrilateral.

    The length of the diagonal BE is  |BE| = {{{sqrt(abs(AE)^2+abs(AB)^2)}}} = {{{sqrt(12^2+1^2)}}} = {{{sqrt(145)}}}.


3.  Therefore, the length of the segment BH is  |BH| = {{{sqrt(abs(BE)^2-abs(HE)^2)}}} = {{{sqrt(145-8^2)}}} = {{{sqrt(81)}}} = 9.


4.  The area of the triangle ABE is {{{(1/2)*abs(AB)*abs(AE)}}} = {{{(1/2)*12*1}}} = 6.

    The area of the triangle BHE is {{{(1/2)*abs(BH)*abs(HE)}}} = {{{(1/2)*9*8}}} = 36.


    Hence, the area of the quadrilateral ABHE is the sum of the areas of the triangles 6 + 36 = 42.


5.  Now the area of the shaded part under the question is 12*8 - 42 = 96-42 = 54.
</pre>

<U>Answer</U>.  The area of the shaded region ABGFEA is 54 square units.


Beautiful problem. Thanks.