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In the adjoining figure,
$AB \parallel EF \parallel CD$, $CD = 12$ cm,
$AB = 7.2$ cm and $DF = 4.8$ cm.
Prove that $\frac{CF}{FB} = \frac{DF}{FA}$.
Also, find the value of $y$, if $x = 4.5$ cm.
$AB \parallel EF \parallel CD$, $CD = 12$ cm,
$AB = 7.2$ cm and $DF = 4.8$ cm.
Prove that $\frac{CF}{FB} = \frac{DF}{FA}$.
Also, find the value of $y$, if $x = 4.5$ cm.
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Ans. In $\triangle ACB, EF \parallel AB$
$\frac{CE}{EA} = \frac{CF}{FB}$ ----- (1) (1 Mark)
In $\triangle ACD, EF \parallel CD$,
$\frac{CE}{EA} = \frac{DF}{FA}$ ----- (2) (1 Mark)
From (1) and (2)
$\frac{CF}{FB} = \frac{DF}{FA}$ (1/2 Mark)
In $\triangle ADC$ and $\triangle AFE$,
$\angle A = \angle A$ (common)
$\angle AFE = \angle ADC$ (corresponding angles)
$\triangle ADC \sim \triangle AFE$ (By AA Similarity) (1
frac{1}{2} Mark)
$\frac{AD}{AF} = \frac{DC}{EF} \implies \frac{y + 4.8}{y} = \frac{12}{4.5}$ (1/2 Mark)
$y = 2.88$ cm (1/2 Mark)
$\frac{CE}{EA} = \frac{CF}{FB}$ ----- (1) (1 Mark)
In $\triangle ACD, EF \parallel CD$,
$\frac{CE}{EA} = \frac{DF}{FA}$ ----- (2) (1 Mark)
From (1) and (2)
$\frac{CF}{FB} = \frac{DF}{FA}$ (1/2 Mark)
In $\triangle ADC$ and $\triangle AFE$,
$\angle A = \angle A$ (common)
$\angle AFE = \angle ADC$ (corresponding angles)
$\triangle ADC \sim \triangle AFE$ (By AA Similarity) (1
frac{1}{2} Mark)
$\frac{AD}{AF} = \frac{DC}{EF} \implies \frac{y + 4.8}{y} = \frac{12}{4.5}$ (1/2 Mark)
$y = 2.88$ cm (1/2 Mark)