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I found a fastest and easier way and I am quite surprised because usually I am not so strong in Quant
Area of the triangle is (B*h)/2 and you can use different "h", there are 3 in every triangle. In this triangle there is QS but also QR and another one that is not drawn (from point P to somewhere in QR where it will be perpendicular) but we don't really care about the last one so it's ok
Area of PQR is (3*4)/2 = 6 (using QR as "h")
The area has to be the same so you can find QS by replacing the data : (5* QS)/2 = 12/5
Sounds ok to me, is there any objection ?
Ps : sorry, just realized crackgmat3 used the same way...I don't delete my post, the explanations are worth it