Question

In the figure given 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 $y$.

Answer 1

Step 1: Find relation between $x$ and $y$

In $△BDC$ and $△BFE,DC\parallel FE$

$\Rightarrow \frac{BC}{BE}=\frac{DC}{FE}=\frac{BD}{BF}$

$\Rightarrow \frac{x}{x+3}=\frac{y}{4.5}=\frac{BD}{BF}\dots \left(1\right)$

Now, In $△FCD$ and $△FAB,DC\parallel AB$

$\Rightarrow \frac{DC}{AB}=\frac{FD}{FB}$

$\Rightarrow \frac{y}{7.5}=\frac{FD}{BF}$




From figure; $FD=BF-BD$

So, $\frac{y}{7.5}=\frac{BF-BD}{BF}$

$\Rightarrow \frac{y}{7.5}=\frac{1-BD}{BF}$

$\Rightarrow \frac{BD}{BF}=\frac{7.5-y}{7.5}\dots \left(2\right)$

Step 2: Finding $x$ and $y$

From (1) and (2), we get;

$\frac{x}{x+3}=\frac{y}{4.5}=\frac{7.5-y}{7.5}$

Or, $\frac{y}{4.5}=\frac{7.5-y}{7.5}$

$\Rightarrow y=2.88cm$

And, $\frac{x}{x+3}=\frac{2.88}{4.5}$

$\Rightarrow x=5.27cm$

Hence, $x=5.27cm$ and $y=2.88cm$.