To work out area of P, Q and R I will use (base*height)/2.
Therefore:
Area of P=(bc)/2
Area of Q=(ac)/2
Area of R=(ab)/2
To prove that P^2+Q^2+R^2=S^2 I will square these areas so that:
Area of P^2=(b^2*c^2)/4
Area of Q^2=(a^2*c^2)/4
Area of R^2=(a^2*b^2)/4
Therefore area S^2=((b^2*c^2)+(a^2*c^2)+(a^2*b^2))/4
and area of S=(((b^2*c^2)+(a^2*c^2)+(a^2*b^2))^(1/2))/2 EQUATION 1
This is the Area of S if P^2+Q^2+R^2=S^2 but to show this is true I need to prove that equation 1 is equal to the area derived by using an area equation on triangle S.
As S is not a right angle triangle using(base*height)/2 is not a suitable way to calculate the area. Therefore I will use Area=(x*y*Sin(C))/2 which is the trigonometry area equation. In this area equation x and y are two of the sides of triangle S. This area also requires me to work out Sin(C). I will do this by using the identity Cos^2(C)+Sin^2(C)=1 by using the cosine rule
Cos(C)=(x^2+y^2-z^2)/(2*x*y), square this and taking this away from 1. As this only gives Sin^2(C) I will work out area^2 by (x^2*y^2*Sin^2(C))/4 and square root this. To figure out Cos(C) I need to have all 3 sides of triangle S which are the hypotenuses of P,Q and R. These are:
Hyp. of P=x=(b^2+c^2)^(1/2)
Hyp. of Q=y=(a^2+c^2)^(1/2)
Hyp. of R=z=(a^2+b^2)^(1/2)
Substituting these into the cosine rule I get:
((b^2+c^2)+(a^2+c^2)-(a^2+b^2))/2*((b^2+c^2)*(a^2+c^2))^(1/2)
This can be simplified to:
(c^2)/((b^2+c^2)*(a^2+c^2))^(1/2)
However I need to square this in order to work out Sin^2(C) and this gives:
(c^4)/(b^2+c^2)*(a^2+c^2)
I can then substitute 1 to give Sin^2(C) which is 1-((c^4)/(b^2+c^2)*(a^2+c^2)) which can also be written as (b^2+c^2)*(a^2+c^2)-(c^4)/(b^2+c^2)*(a^2+c^2).
This can be then substituted into Area of S^2=(x^2*y^2*Sin^2(C))/4 which gives:
((b^2*c^2)+(a^2*c^2)+(a^2*b^2))/4
Therefore Area of S = (((b^2*c^2)+(a^2*c^2)+(a^2*b^2))^(1/2))/2
This is the same as the equation derived earlier. This proves that R^2+Q^2+P^2=S^2.