In the figure given below, AB ‖ EF ‖ CD. If AB = 22.5 cm, EP = 7.5 cm, PC = 15 cm and DC = 27 cm.

Calculate: (i) EF (ii) AC

(i) In ∆PCD and ∆PEF,

∠CPD = ∠EPF [Vertically opposite angles are equal]

∠DCP = ∠FEP [Alternate angles are equal]

∴ ∆PCD ~ ∆PEF [By AA]

Since corresponding sides of similar triangle are proportional to each other.

$\Rightarrow \dfrac{CD}{EF} = \dfrac{PC}{EP} \\[1em] \Rightarrow \dfrac{27}{EF} = \dfrac{15}{7.5} \\[1em] \Rightarrow EF = \dfrac{27 \times 7.5}{15} \\[1em] \Rightarrow EF = 13.5 \text{ cm}.$

Hence, EF = 13.5 cm.

(ii) In ∆CEF and ∆CAB

∠FCE = ∠BCA [Common angles]

∠CFE = ∠CBA [As FE || BA, corresponding angles are equal]

∴ ∆CEF ~ ∆CAB [By AA]

Since corresponding sides of similar triangle are proportional to each other.

$\Rightarrow \dfrac{EC}{AC} = \dfrac{EF}{AB} \\[1em] \Rightarrow \dfrac{PC + PE}{AC} = \dfrac{13.5}{22.5} \\[1em] \Rightarrow \dfrac{15 + 7.5}{AC} = \dfrac{13.5}{22.5} \\[1em] \Rightarrow \dfrac{22.5}{AC} = \dfrac{13.5}{22.5} \\[1em] \Rightarrow AC = \dfrac{22.5 \times 22.5}{13.5} \\[1em] \Rightarrow AC = 37.5 \text{ cm}.$

Hence, AC = 37.5 cm.