Question

In the following figure, ABCD to a trapezium with AB || DC. If AB = 9 cm, DC = 18 cm, CF = 13.5 cm, AP = 6 cm and BE = 15 cm.

Calculate:

1. EC
2. AF
3. PE

Answer 1

i. In ΔAEB and ΔFEC,

∠AEB = ∠FEC  ...(Vertically opposite angles)

∠BAE = ∠CFE  ...(Since AB || DC)

ΔAEB ∼ ΔAFEC  ...(AA criterion for similarity)

`=> (AE)/(FE) = (BE)/(CE) = (AB)/(FC)`

`=> 15/(CE) = 9/13.5`

`=>` CE = 22.5 cm

ii. In ΔAPB and ΔFPD 

∠APB = ∠FPD  ...(Vertically opposite angles)

∠BAP = ∠DFP  ...(Since AB || DF)

ΔAPB ∼ ΔFPD  ...(AA criterion for similarity)

`=> (AP)/(FP) = (AB)/(FD)`

`=> 6/(FP) = 9/31.5`

`=>` FP = 21 cm

So AF = AP + PF

= 6 + 21

= 27 cm

iii. Given: ABCD is trapezium, AB || DC

AB = 9 cm, DC = 18 cm, CF = 13.5 cm, AP = 6 cm, BE = 15 cm

Consider ΔAPB and ΔFPD,

`=>` ∠APB = ∠FPD   ...(Vertically opposite angles)

`=>` ∠BAP = ∠DFP   ...(Since AB || DF)

ΔAPB ~ ΔFPD     ...(AA criterion for similarity)

`=> (AP)/(FP) = (AB)/(FD)`

`=> 6/(FP) = 9/31.5`

`=>` FP = 21 cm

So, AF = AP + PF

= 6 + 21

= 27 cm

In ΔAEB and ΔFEC,

∠AEB = ∠FEC  ...(Vertically opposite angles)

∠BAF = ∠CFE    ...(Since AB || DC)

ΔAEB ~ ΔFEC    ...(AA criterion for Similarity)

`=> (AF)/(FE) = (AB)/(FC)`

`=> (AP + PE)/(FE) = 9/13.5`

`=> (AP + PE)/(FE) = 90/135`

`=> (6 + PE)/(FE) = 30/45 = 2/3`

`=>` 3(6 + PE) = 2FE

`=> (3(6 + PE))/2 = FE`

Now, PF = PE + EF

`21 = PE + (3 (6 + PE))/2`

`21 = (2PE + 18 + 3PE)/2`

42 = 5PE + 18

42 – 18 = 5PE

24 = 5PE

`24/5` = PE

PE = 4.8