ABC is a triangle with AB = 12, AC = 14 and BC = 16.
D is the midpoint of AB.
The vertex C is joined with D forming the crease EF with E on AC and F on BC.
Find the area of the quadrilateral BDEF
My plan is to find the areas of ΔADE and ΔCEF and subtract them from ΔABC.
First, by Heron's formula, Area of ΔABC=21√15.
By law of cosines:
cos(BAC)=1/4, sin(BAC)=√(15)/4
then by law of cosines again
CD=√(190), CG=GD=√(190)/2
By law of sines
sin(ACD)=3√(114)/76, cos(ACD)=5√(190)/76
By right triangle trig on ΔCGE,
CE=38/5,
ED=38/5 because CGE and DGE are congruent
AE=1438/5=32/5
Area ΔAED = (1/2)(6)(32/5)(√(15)/4)=24√(15)/5
By law of cosines
cos(BCD)=41√(190)/608
By right triangle trig on ΔCGF,
CF=304/41
By law of cosines
cos(ABC)=11/16, sin(ABC)=3√(15)/16
Area ΔEFC=(1/2)(38/5)(304/41)(3√(15)/16)=1083√(15)/205
Finally
Area BDEF = 21√(15)  24√(15)/5  1083√(15)/205 =
2238√(15)/205 ≈ 42.28
This seems pretty complicated but I don't see how it could be done more simply.

Posted by Jer
on 20160410 16:42:22 