In figure below, we have $AB \parallel CD \parallel EF$. If $AB=6\ cm, CD=x\ cm, EF=10\ cm, BD=4\ cm$ and $DE=y\ cm$, calculate the values of $x$ and $y$."

Given:

$AB \parallel CD \parallel EF$.

$AB=6\ cm, CD=x\ cm, EF=10\ cm, BD=4\ cm$ and $DE=y\ cm$.
To do:

We have to calculate the values of $x$ and $y$.

Solution:

In $\triangle ECD$ and $\triangle EAB$,

$\angle ECD=\angle EAB$   (Corresponding angles)

$\angle CED=\angle AEB$   (Common angle)

Therefore,

$\triangle ECD \sim\ \triangle EAB$  (By AA similarity)

This implies,

$\frac{CD}{AB}=\frac{CE}{AE}$

$\frac{x}{6}=\frac{CE}{AE}$.....(i)

Similarly,

$\frac{CD}{AB}=\frac{DE}{BE}$

$\frac{x}{6}=\frac{y}{y+4}$......(ii)

In $\triangle ADC$ and $\triangle AFE$,

$\angle ACD=\angle AEF$   (Corresponding angles)

$\angle DAC=\angle FAE$   (Common angle)

Therefore,

$\triangle ADC \sim\ \triangle AFE$  (By AA similarity)

This implies,

$\frac{CD}{EF}=\frac{AC}{AE}$

$\frac{x}{10}=\frac{AC}{AE}$.....(iii)

Adding equations (i) and (iii), we get,

$\frac{x}{6}+\frac{x}{10}=\frac{CE}{AE}+\frac{AC}{AE}$

$\frac{5x+3x}{30}=\frac{AC+CE}{AE}$

$\frac{8x}{30}=\frac{AE}{AE}$   ($AC+CE=AE$)

$\frac{4x}{15}=1$

$4x=15$

$x=\frac{15}{4}$

$x=3.75\ cm$

From equation (ii),

$\frac{\frac{15}{4}}{6}=\frac{y}{y+4}$

$\frac{15}{24}=\frac{y}{y+4}$

$15(y+4)=y(24)$

$15y+60=24y$

$24y-15y=60$

$9y=60$

$y=\frac{60}{9}$

$y=6.67\ cm$

The values of $x$ and $y$ are $3.75\ cm$ and $6.67\ cm$ respectively.

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Updated on: 10-Oct-2022

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