In AABC, P is a point on AB such that PB: AP = 3: 4 and

Solution

△ ABC with P on AB such that PB: AP = 3: 4. PQ || AC. AR ⊥ PC and QS ⊥ PC. QS = 9 cm. Proportional Segments: Given – PB: AP= 3: 4, let PB = 3k and AP = 4k. Therefore, AB = PB + AP=3k =4k = 7k. Similar Triangles: Since PQ || AC, triangles △ APQ and △ APC are similar. This implies the ratio of their corresponding sides is. PB/AB=3/7 Area of △PBQ/△ABC =9/49 Ratio of △APC and △QPC- =½ ×PC×AR: ½×PC×QS =AR: QS △APC/△QPC =AR/QS            … (1) Area of △PBQ=9 and △QPC =12 Now Area of △APC =△ABC –(△PBQ+△QPC) =49-(9+12) =28. From Eq-(1) 28/12 =AR/9 AR =21cm.