In the figure, given below, A B, C D and E F are parallel lines. … - Vidyadip

Question

In the figure, given below, $A B, C D$ and $E F$ are parallel lines. Given $A B=7.5 cm, D C=y cm, E F=4.5 cm, B C=x cm$ and $C E=3 . cm$, calculate the values of $x$ and $y$.

✓

Answer

In $\triangle BEF, DC || EF$
$\Rightarrow \frac{B D}{D F}=\frac{B C}{C E}$
$\Rightarrow \frac{B D}{D F}=\frac{ x }{3}$
So, $BD = x$ and $DF = 3$
In $\triangle AFB, DC || AB.$
$\Rightarrow \frac{F D}{C D}=\frac{F B}{A B}$
$\Rightarrow \frac{F D}{C D}=\frac{F D+D B}{A B}$
$\Rightarrow \frac{3}{y}=\frac{ x +3}{7.5} \ldots . .$. (i)
In $\triangle BEF, DC || EF.$
$\Rightarrow \frac{B C}{C D}=\frac{B E}{E F}$
$\Rightarrow \frac{B C}{C D}=\frac{B C+C E}{E F}$
$\Rightarrow \frac{ x }{ y }=\frac{ x +3}{4.5}$
$\Rightarrow y=\frac{4.5 x }{ x +3} \ldots . . \text { (ii) }$
Substituting $(ii)$ in $(i)$, we get
$\frac{3}{\frac{4.5 x}{x+3}}=\frac{x+3}{7.5}$
$\Rightarrow \frac{3 x+9}{4.5 x}=\frac{x+3}{7.5}$
$\Rightarrow 22.5x + 67.5 = 4.5x^2+ 13.5x$
$\Rightarrow 4.5x^2+ 13.5x - 22.5x - 67.5 = 0$
$\Rightarrow x^2- 2x - 15 = 0$
$\Rightarrow (x - 5)(x + 3) = 0$
So, $x=5$ and $x=-3$
Since side of a triangle cannot be negative, $x=5$
Substituting this value in (ii), we get
$y=\frac{4.5(5)}{x+3}=2.8125$
Hence, $x=5$ and $y=2.8125$

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