Abcd is a trapezium in which ab parallel to dc and p and q are points on ad and bc

ABCD is a trapezium in which AB parallel DC and P and Q are points on AD and BC respectively such that PQ parallel DC . if PD = 12 cm , BQ = 42 cm ,and QC = 18 cm , find AD.(experts please give the answers with same numbers which is provided in question don’t change the numbers and don’t provide any link.) Join BD that intersects PQ at L . Since , LQ ∥ DC and BD is a transversal , then ∠ BLQ = ∠ BDC Corresponding angles Since , LQ ∥ DC and BC is a transversal , then ∠ BQL = ∠ BCD Corresponding angles In ∆ BLQ and ∆ BDC ∠ BLQ = ∠ BDC Proved above ∠ BQL = ∠ BCD Proved above ⇒ ∆ BLQ ~ ∆ BDC AA ⇒ BL BD = LQ DC = BQ BC Corresponding sides of similar ∆ ' s are proportional ⇒ BL BD = BQ BC ⇒ BD BL = BC BQ ⇒ BD BL - 1 = BC BQ - 1 ⇒ BD - BL BL = BC - BQ BQ ⇒ DL BL = CQ BQ . . . . . 1 Since , PL ∥ AB and DB is a transversal , then ∠ DLP = ∠ DBA Corresponding angles Since , PL ∥ AB and DA is a transversal , then ∠ DPL = ∠ DAB Corresponding angles In ∆ DLP and ∆ DBA ∠ DLP = ∠ DBA Proved above ∠ DPL = ∠ DAB Proved above ⇒ ∆ DLP ~ ∆ DBA AA ⇒ DL DB = LP BA = DP DA Corresponding sides of similar ∆ ' s are proportional ⇒ DL DB = DP DA ⇒ DB DL = DA DP ⇒ DB DL - 1 = DA DP - 1 ⇒ DB - DL DL = DA - DP DP ⇒ BL DL = AP DP ⇒ DL BL = DP AP . . . . . 2 From 1 and 2 , we get CQ BQ = DP AP ⇒ 18 42 = 12 AP ⇒ AP = 42 × 12 18 = 28 cm Now , AD = AP + DP = 28 + 12 = 40 cm

Given: $ABCD$ is a trapezium in which $AB \parallel DC$ and $DC = 40\ cm$ and $AB = 60\ cm$. $X$ and $Y$ are respectively the mid-points of $AD$ and $BC$. To do: We have to prove that (i) $XY = 50\ cm$. (ii) $DCYX$ is a trapezium. (iii) $ar(trap.\ DCYX) = \frac)$ Hence proved.

The required figure is shown below (i) From ΔPED and ΔABP, PD = AP...[P is the mid-point of AD] ∠DPE = ∠APB ....[Opposite angle] ∠PED = ∠PBA ...[AB || CE] ∴ΔPED ≅ΔABP...[ASA postulate] ∴ EP = BP (ii) In Δ ECB, P is a mid point of BE and Q is a mid point of BC ∴ PQ || CE ...(i) (by mid point theorem) and CE || AB ... (ii) From equation (i) and (ii) PQ || AB Hence proved.