Question

In the figure given, AB, CD, and EF are parallel lines.
B = 9 cm, DC = y cm, EF = 4.5 cm, BC = x cm, and CE = 3 cm. Find the values of 'x' and 'y'.

Answer 1

In $\Delta BFE\&BDC$
$\angle B=\angle B$ (common angle)
$\angle BDC=\angle BFE$ (corresponding angle)
$\therefore \Delta BFE\sim \Delta BDC$ (AA similarity criterion)

So, $\frac{BC}{BE}=\frac{CD}{EF}=\frac{BD}{BF}$ (Ratio of corresponding sides are equal in similar triangles)
$\frac{x}{x+3}=\frac{y}{4.5}=\frac{BD}{BF}$ ------ (1)

In $\Delta FDC\&\Delta FBA$
$\angle F=\angle F$ (common)
$\angle FDC=\angle FBA$ (corresponding angle)
$\therefore \angle FDC\sim \Delta FBA$ (AA similarity criterion)

So, $\frac{FD}{BF}=\frac{y}{7.5}$ ------- (2)

From (1) we have $\frac{BD}{BF}=\frac{y}{4.5}$ ------- (3)

Adding (2) and (3), we get
$\frac{FD}{BF}+\frac{BD}{BF}=\frac{y}{7.5}+\frac{y}{4.5}\frac{FD+BD}{BF}=y(\frac{1}{7.5}+\frac{1}{4.5})\frac{BF}{BF}=y\left(\frac{4.5+7.5}{7.5\times 4.5}\right)y=\frac{7.5\times 4.5}{12}=\frac{33.75}{12}=\frac{45}{16}$ cm

From (1) we have,

$\Rightarrow \frac{x}{x+3}=\frac{y}{4.5}\Rightarrow \frac{x}{x+3}=\frac{45}{16\times 4.5}\Rightarrow \frac{x}{x+3}=\frac{5}{8}\Rightarrow 8x=5x+15\Rightarrow x=5$ cm