CPhill

https://web2.0calc.com/questions/polygon_72

https://web2.0calc.com/questions/circles_146

https://web2.0calc.com/questions/equilateral

3 - 3 - 1 - 1

3 - 1 - 1 - 1 - 1 - 1

1 -1- 1 -1- 1 -1 - 1 -1

PQ  = sqrt [ 2^2 + 3^2 ] = sqrt [ 13]

AQ  = sqrt [1^2 + 4^2 ] sqrt [17]

PA  = sqrt  [2^2 + 4^2] = sqrt [ 20]  

Law of Cosines

PQ^2  = AQ^2 + PA^2  - 2 ( AQ * PA) (cos PAQ)

13 = 17 + 20 - 2( sqrt 340) * cos (PAQ)

[ 13 - 17 -20 ] / [ -2sqrt 340 ]  = cos (PAQ) = 6/sqrt 85

sin PAQ =   sqrt [ 85 - 6^2 ] / sqrt 85  = sqrt [ 49] / sqrt 85  =  7 / sqrt 85

ABC  ≈ XYZ

BC / AB = YZ / XY

8 / 14  = YZ / 18

4/ 7   = YZ / 18

YZ =  [ 18 / 4 ] / 7  =  32 / 7

Hey...you got it  !!!!

Here's the same problem with a side of 12

https://web2.0calc.com/questions/square_34   (proyaop's answer )

See if you can follow the reasoning with your problem

Hint :  I believe that your "PR"   will be  (10/12)  of what was found in this link 

When we have this situation..

Area =   (1/2) ( product of sides between the included angle) * sine (included angle)

See what you come up with....if you get stuck....let me know

British Flag Theorem

AX^2  + CX^2 = BX^2  + DX^2

Can you take it  from  here ???