Question

In the given figure $AB\parallel EF\parallel DC$; AB = 67.5 cm. DC = 40.5 cm and AE = 52.5 cm. Find the length of EF.

• A. $8\frac{5}{16}\text{cm}$.
• B. $20\frac{5}{16}\text{cm}$.
• C. $15\frac{5}{16}\text{cm}$.
• D. $25\frac{5}{16}\text{cm}$.

Answer 1

The correct option is D

$25\frac{5}{16}\text{cm}$.

(i) In the figure

$\because AB\parallel EF\parallel DC$

$\therefore$ There are three pairs of similar triangles

(i) $\Delta AEB\sim \Delta DEC$

(ii) $\Delta ABC\sim \Delta EFC$

(iii) $\Delta AEB\sim \Delta DEC$

$\therefore \frac{AE}{EC}=\frac{BE}{ED}=\frac{AB}{DC}$

But AB = 67.5 cm, DC = 40.5 cm and AE = 52.5 cm

$\therefore \frac{52.5}{EC}=\frac{67.5}{40.5}=EC\times 67.5=52.5\times 40.5$

$EC=\frac{52.5\times 40.5}{67.5}=31.5\text{cm}$

In $\Delta ABC$, $EF\parallel AB$

$\therefore \frac{AC}{EC}=\frac{AB}{EF}$

$\frac{AE+EC}{EC}=\frac{AB}{EF}=\frac{52.5+31.5}{31.5}=\frac{67.5}{EF}$

$\Rightarrow \frac{84}{31.5}=\frac{67.5}{EF}\Rightarrow 84\times EF=67.5\times 31.5$

$\therefore EF=\frac{67.5\times 31.5}{84}=\frac{675\times 315}{10\times 84\times 10}$

$=\frac{405}{16}=25\frac{5}{16}\text{cm}$.