Let angle ABD be expressed as T.
Then angle BAD is pi-2T and angle ADC is pi-T.
Then the area of ABD is (1/2)*1*1*sin(pi-2T) and the area of ACD is (1/2)*1*1*sin(pi-T).
The area(E) of the whole triangle is:
E = (1/2)*1*1*sin(pi-2T) + (1/2)*1*1*sin(pi-2T)
E simplifies to E = (sin(2T)+sin(T))/2
The value of T which maximizes the area is 0, pi/2, or a relative maxima of E in the range [0,pi/2].
For T=0, the area is 0. For T=pi/2, the area is 1/2. To find the relative maxima, differentate E and solve dE/dT=0.
dE/dT = (2*cos(2T)+cos(T))/2 = 0
Simplifying yields 4(cos(T))^2 + cos(T) - 2 = 0
Using the quadratic formula gives cos(T)=.59307 or cos(T)=-.84307.
The only value of T in [0,pi/2] which is a solution is T=.93593 radians (from cos(T)=.59307).
With T=.93593 radians (= 53.625 degrees), the maximum area is .88009 |
