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 :

1. EF
2. AC

Answer 1

In the given figure,

AB || EF || CD

AB = 22.5 cm, EP = 7.5 cm

PC = 15 cm and DC = 27 cm

To calculate:

1. EF
2. AC

In ΔDCP and ΔPEF

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

∠DCP = ∠PEF   ...(Alternate angles)

∴ ΔDCP ∼ ΔPEF  ...(AA axiom)

∴ ${\displaystyle \frac{DC}{EF}=\frac{CP}{PE}}$

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

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

= ${\displaystyle \frac{27}{2}}$

= 13.5 cm

Similarly in ΔABC, EF || AB

∴ ΔCEF ∼ ΔACB

∴ ${\displaystyle \frac{CE}{AC}=\frac{EF}{AB}}$

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

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

= ${\displaystyle \frac{225\times 225}{135\times 10\times 10}}$

= ${\displaystyle \frac{375}{10}}$

= 37.5 cm

Thus AC = 37.5 cm