Insane Area Chasing!

As noted by two friends below, ABC is 3-4-5 right angle at B.
So SinA=4/5, SinC=3/5.
So areas of triangles at ,
$A \ is \ 1/2*4*3*4/5=6,..........B \ is \ 1/2*2*4*4/5=16/5,...........C \ is \ 1/2*6*5*3/5=9. \\ Area \ ABC=1/2*6*8=24. \\. So \ the \ area \ of\ PQR=24-6-16/5-9=**5.8**.$

one thing to note here is that the bigger triangle ABC is a right angle triangle (6,8,10) and hence all the included angles can be found out very easily. Now considering all the small triangles i.e. PRC, PQB, QAR, using the formula A= ( Sin(included angle)*product of containing sides)/2 , the areas are calculated and by subtracting them from the area of ABC we get the required answer.

First,note that its sides are(6,8,10) which means its a right triangle.Now B=90,and sinC=(6/10) or C=37 and same way A=53.Using cosine law we can find out the sides of the triangle PQR. ie QR^2=2^2+4^2-2 4 2*cos(53)=10.37 or QR=3.22.Same way PR is 3.616 and QP=5(we can get it by pythagorean theorem too).We know all the sides of the triangle so we can compute the area with Heron's formula which becomes 5.8