ABCD of a trapezium in which AB||CD and P.Q are Point on ABC respectively such that PQ||DC if PD=18cm BQ=35cm and QC=15cm, find AD[see figure]
Answers
Join BD that intersects PQ at L.Since, LQ∥DC and BD is a transversal, then∠BLQ = ∠BDC Corresponding anglesSince, LQ∥DC and BD is a transversal, then∠BQL = ∠BCD Corresponding anglesIn ∆BLQ and ∆BDC∠BLQ = ∠BDC Proved above∠BQL = ∠BCD Proved above⇒∆BLQ ~ ∆BDC AA⇒BLBD = LQDC = BQBC Corresponding sides of similar ∆'s are proportional⇒BLBD = BQBC ⇒BDBL = BCBQ⇒BDBL - 1= BCBQ - 1⇒BD - BLBL = BC - BQBQ⇒DLBL = CQBQ .....1Since, PL∥AB and DB is a transversal, then∠DLP = ∠DBA Corresponding anglesSince, PL∥AB and DB is a transversal, then∠DPL = ∠DAB Corresponding anglesIn ∆DLP and ∆DBA∠DLP = ∠DBA Proved above∠DPL = ∠DAB Proved above⇒∆DLP ~ ∆DBA AA⇒DLDB = LPBA = DPDA Corresponding sides of similar ∆'s are proportional⇒DLDB = DPDA ⇒DBDL = DADP⇒DBDL-1 = DADP-1⇒DB - DLDL = DA-DPDP⇒BLDL = APDP⇒DLBL = DPAP .....2From 1 and 2, we getCQBQ = DPAP⇒1842 = 12AP⇒AP = 42 × 1218 = 28 cmNow, AD = AP + DP = 28 + 12 = 40 cm
Answer:
Join BD that intersects PQ at L.Since, LQ∥DC and BD is a transversal, then∠BLQ = ∠BDC Corresponding anglesSince, LQ∥DC and BD is a transversal, then∠BQL = ∠BCD Corresponding anglesIn ∆BLQ and ∆BDC∠BLQ = ∠BDC Proved above∠BQL = ∠BCD Proved above⇒∆BLQ ~ ∆BDC AA⇒BLBD = LQDC = BQBC Corresponding sides of similar ∆'s are proportional⇒BLBD = BQBC ⇒BDBL = BCBQ⇒BDBL - 1= BCBQ - 1⇒BD - BLBL = BC - BQBQ⇒DLBL = CQBQ .....1Since, PL∥AB and DB is a transversal, then∠DLP = ∠DBA Corresponding anglesSince, PL∥AB and DB is a transversal, then∠DPL = ∠DAB Corresponding anglesIn ∆DLP and ∆DBA∠DLP = ∠DBA Proved above∠DPL = ∠DAB Proved above⇒∆DLP ~ ∆DBA AA⇒DLDB = LPBA = DPDA Corresponding sides of similar ∆'s are proportional⇒DLDB = DPDA ⇒DBDL = DADP⇒DBDL-1 = DADP-1⇒DB - DLDL = DA-DPDP⇒BLDL = APDP⇒DLBL = DPAP .....2From 1 and 2, we getCQBQ = DPAP⇒1842 = 12AP⇒AP = 42 × 1218 = 28 cmNow, AD = AP + DP = 28 + 12 = 40 cm