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

Calculate :

(i) EC (ii) AF (iii) PE

(i) In ΔAEB and ΔFEC,

∠AEB = ∠FEC [Vertically opposite angles are equal]

∠BAE = ∠CFE [Alternate angles are equal]

∴ ∆AEB ~ ∆FEC [By AA]

Since, corresponding sides of similar triangles are proportional.

$\Rightarrow \dfrac{BE}{EC} = \dfrac{AB}{FC} \\[1em] \Rightarrow \dfrac{15}{EC} = \dfrac{9}{13.5} \\[1em] \Rightarrow EC = \dfrac{15 \times 13.5}{9} = 22.5\text{ cm}$

Hence, EC = 22.5 cm.

(ii) In ΔAPB and ΔFPD,

∠APB = ∠FPD [Vertically opposite angles are equal]

∠BAP = ∠DFP [Alternate angles are equal]

∴ ∆APB ~ ∆FPD [By AA]

Since, corresponding sides of similar triangles are proportional.

$\Rightarrow \dfrac{AP}{FP} = \dfrac{AB}{FD} \\[1em] \Rightarrow \dfrac{AP}{FP} = \dfrac{AB}{DC + CF} \\[1em] \Rightarrow \dfrac{6}{FP} = \dfrac{9}{18 + 13.5} \\[1em] \Rightarrow \dfrac{6}{FP} = \dfrac{9}{31.5} \\[1em] \Rightarrow FP = \dfrac{6 \times 31.5}{9} \\[1em] \Rightarrow FP = \dfrac{63}{3} = 21\text{ cm}.$

From figure,

AF = AP + FP = 6 + 21 = 27 cm.

Hence, AF = 27 cm.

(iii) We already have, ∆AEB ~ ∆FEC

So,

$\Rightarrow \dfrac{AE}{FE} = \dfrac{AB}{FC} \\[1em] \Rightarrow \dfrac{AE}{FE} = \dfrac{9}{13.5} \\[1em] \Rightarrow \dfrac{(AF – EF)}{ FE} = \dfrac{9}{13.5} \\[1em] \Rightarrow \dfrac{AF}{EF} – 1 = \dfrac{9}{13.5} \\[1em] \Rightarrow \dfrac{27}{EF} = \dfrac{9}{13.5} + 1 \\[1em] \Rightarrow \dfrac{27}{EF} = \dfrac{22.5}{13.5} \\[1em] \Rightarrow EF = \dfrac{(27 \times 13.5)}{22.5} = 16.2 \text{ cm}.$

From figure,

PE = PF – EF = 21 – 16.2 = 4.8 cm

Hence, PE = 4.8 cm