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In APQR, 2Q = 90°,Qs L PR,QS=12, SR= 8, Find PS. |
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Answer» ΔPQR in which ∠Q = 90°, QS ⊥ PR and PQ = 6 cm, PS = 4 cm In ΔSQP and ΔSRQ,ANSWERIN triangles PQS and PQR∠P is common to both.and ∠PSQ=∠PQRTriangles PQS and PQR are equiangular∴ PQPS = QRQS or, PS= QRPQ⋅QS ...(i)Again, triangles QRS and PQR are equiangular∴ QRSR = PQQS or, SR= PQQS⋅QR ...(ii)From eqns. (i) and (ii)SRPS = QRPQ⋅QS ⋅ QR⋅QSPQ = QR 2 PQ 2 |
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