FatemehAsgarinejad wrote:

As PRQ is right side we have:

PR^2 + PQ^2 = RQ^2

(PS +SR)^2 + 12^2 = 20^2

(PS +SR) = 16

We know That in a triangle, the summation of each two sides is more than the third one, so:

12 < PS + QS

20 < SR + QS so 20- 8 < (SR-8) + QS

From above two equations, we recognize that SR is at least 8 more than PS. So SR is more than PS.

NOT SURE IT'S CORRECT OR NOT

how can you make sure "20- 8 < (SR-8) + QS" ?

I think the ans is D.

if we cut PR into half >>> PS=SR=8

then QS would be around 14. which is applicable for a right trianle with sides 12-8-14.XX