Triangle ABC has AB =20, BC =34 and CA =42. Point P lies on the circle with diameter AB. The greatest possible area of triangle APC is R, then which of the following options is (are) CORRECT?

Area of triangle APC is greatest
$\Rightarrow$ Height of the triangle is maximum
Maximum height of triangle APC = PQ = MP + MQ
M is mid point of AB $\Rightarrow$ Centre of circle
MP = Radius of circle =  $\frac{1}{2}\left(20\right)=10$
MQ is largest  $\Rightarrow \frac{MQ}{AM}=\frac{BI}{AB}\Rightarrow MQ=\frac{1}{2}{h}_B$
BT = Altitude of triangle through vertex B to side AC
$h_B=\frac{2\Delta }{b}=\frac{2\Delta }{AC}$ 
$\Delta =\sqrt{\left(48\right)\left(28\right)\left(14\right)\left(6\right)}=336$ 
 $h_B=\frac{2\left(336\right)}{42}=16$
 $MQ=\frac{1}{2}{h}_B=8$
PQ = MP + MQ = 10 + 8 = 18
Maximum area of  $\Delta \mathcal{l}eAPC=\frac{1}{2}\times 42\times 18=\left(21\right)\left(18\right)=378$

$R=378=2^1{3}^3{.7}^1$
Number of positive integral divisors = (1 + 1) (3 + 1) (1 + 1)  = (2) (4) (2) = 16