Question

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

• A. x$=$ 1 cm and $y=2\frac{13}{16}$ cm
• B. x$=$ 5 cm and $y=2\frac{13}{16}$ cm
• C. x$=$ 7 cm and $y=2\frac{13}{16}$ cm
• D. x$=$ 2 cm and $y=2\frac{13}{16}$ cm

Answer 1

Answer: (B)

x$=$ 5 cm 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)
$\frac{x}{x+3}=\frac{BD}{BF}=\frac{y}{4.5}$ ..(I)
Similarly, $△FCD\sim △FAB$
Thus, $\frac{FC}{FA}=\frac{CD}{AB}=\frac{FD}{FB}$
$\frac{y}{7.5}=\frac{DF}{BF}$ (II)
From I and II,
$BD=\frac{y}{4.5}BF$ and $DF=\frac{y}{7.5}BF$
$BD+DF=BF$
$\frac{y}{4.5}BF+\frac{y}{7.5}BF=BF$
$7.5y+4.5y=7.5\times 4.5$
$y=\frac{45}{16}$
Also, $\frac{x}{x+3}=\frac{y}{4.5}$
$\frac{x}{x+3}=\frac{45}{16\times 4.5}$
$\frac{x}{x+3}=\frac{45}{72}$
$72x=45x+135$
$27x=135$
$x=5cm$