Question

In the figure below $AB$, $CD$ and $EF$ are parallel lines. Given $AB=7.5cm$ , $DC=ycm$. $EF=4.5cm$, $BC=xcm$ and $CE=3cm$. Calculate the values of $x$ and $y$.

• A. $x=5cm$ and $y=2\frac{13}{16}cm$
• B. $x=5cm$ and $y=1\frac{13}{16}cm$
• C. $x=5cm$ and $y=3\frac{13}{16}cm$
• D. $x=5cm$ and $y=4\frac{13}{16}cm$

Answer 1

The correct option is A $x=5cm$ and $y=2\frac{13}{16}cm$

In $△BCD$ and $△BEF$,

$\angle DBC=\angle FBE$ (Common angle)

$\angle BCD=\angle BEF$ (Corresponding angles of parallel lines CD and EF)

$\angle BDC=\angle BFE$ (Corresponding angles of parallel lines CD and EF)

Thus, $△BCD\sim △BEF$ (AAA rule)

Hence, $\frac{BC}{BE}=\frac{BD}{BF}=\frac{CD}{EF}$ (Corresponding sides)

$\Rightarrow \frac{x}{x+3}=\frac{BD}{BF}=\frac{y}{4.5}$ ..(I)
Similarly, $△FCD\sim △FAB$

Thus, $\frac{FC}{FA}=\frac{CD}{AB}=\frac{FD}{FB}$

$\Rightarrow \frac{y}{7.5}=\frac{DF}{BF}$ (II)

From I and II,

$\Rightarrow BD=\frac{y}{4.5}BF$ and $DF=\frac{y}{7.5}BF$

$\Rightarrow BD+DF=BF$

$\Rightarrow \frac{y}{4.5}BF+\frac{y}{7.5}BF=BF$

$\Rightarrow 7.5y+4.5y=7.5\times 4.5$

$\Rightarrow y=\frac{45}{16}$
Also, $\frac{x}{x+3}=\frac{y}{4.5}$

$\Rightarrow \frac{x}{x+3}=\frac{45}{16\times 4.5}$

$\Rightarrow \frac{x}{x+3}=\frac{45}{72}$

$\Rightarrow 72x=45x+135$

$\Rightarrow 27x=135$

$\Rightarrow x=5cm$