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=14-38/5=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 2016-04-10 16:42:22