Area of a rhombus or kite

[BMAD]

Suppose we have a ten inch line segment. Draw two isosceles triangles with legs of 3inches such that each begin at the end of the line and their hypotenuse is opposite of the base line. Extend the hypotenuse lines such that they cross above the middle of the line segment. Draw a line from the height of each triangle to the bottom end of the other triangle (the acute angle). At the top a four-sided shape is formed. What is the area of this shape?  Solve it without calculus. 

[rocdocmac]

No reply yet? Using basics (Pythagoras, sin or cos rule, etc.) - someone can later spell it out in "calculus"! Basically ...

 

Spoiler

HF = 3
CF = 7
CH = 58^0.5 = 7.615773
Angle x = 23.2 deg
CE = 5
OE = 2.142
AO  = 2.858
JH = 2
Area [ABOH] = JH *AO = 5.716

 

Edited November 14, 2016 by bonanova
Spoiler

[BMAD]

https://drive.google.com/file/d/0B62_deMDDaNveTlUSWk5Q0FteE0/view?usp=drivesdk

[plainglazed]

Spoiler

7 3/7?  

Strike that - 5 5/7 instead.  

Top half of the rhombus/kite = 4  

From similar triangles: 3/7=x/3 or x=9/7 and bottom half of rhombus/kite = ((3-x)/2)*2 or 1 5/7.

EDIT:  yeah, really should not post at 4:00 a.m. Have added to the spoiler above. 

[rocdocmac]

 

Spoiler

 

Edited November 14, 2016 by bonanova
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[bonanova]

I spent way too much time trying to figure out the calculus ban. Like there a catch and it should be obvious at some point. If there was a catch I didn't see it.
 

Spoiler

The point coordinates are
A: (5,5) B: (0,0) C: (10,0) D: (3,3) E: (7,3) F: (5, 15/7) G: (3,0) H: (7,0) I: (3,9/7) J: (7,9/7)

ABC = 25
FBC = 75/7
BDI = ECJ = BDG - BIG = 9/2 - (1/2)(27/7) = 18/7
DFI = FEJ = (1/2)(2)(3-9/7) = 12/7

AEFD = ABC - (FCB + 2xBDI + 2xDFI) = 40/7 ~ 5.71428