Question

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 : AC

 

Answer 1

In ΔPCD and ΔPEF,

∠CPD = ∠EPF .....(vertically opposite angles)

∠DCE = ∠FEP ...(since DC || EF)

ΔPCD ∼ ΔPEF ...(AA criterion for similarity)

${\displaystyle \Rightarrow \frac{27}{EF}=\frac{15}{7.5}}$

${\displaystyle \Rightarrow }$ EF = 13.5 cm

Since EF || AB, ΔCEF ∼ ΔCAB

${\displaystyle \Rightarrow \frac{EC}{AC}=\frac{EF}{AB}}$

${\displaystyle \Rightarrow \frac{22.5}{AC}=\frac{13.5}{22.5}}$

${\displaystyle \Rightarrow }$ AC = 37.5 cm