Question

In the given figure , AB || CD || EF given AB = 7.5 cm , DC =y cm , EF = 4.5 cm , BC = x cm . Calculate the values of x and y. .​

Answer 1

$\huge\mathbb{SOLUTION:-}$

$According \:to$

$Thales\: theorem$

∆CDC ∼ ∆ABE

➼ $\frac{cd}{ab} = \frac{de}{be}$

➼ $\frac{y}{7.5} = \frac{de}{be}$

➼ $1 - \frac{y}{7.5} = 1 - \frac{de}{be}$

➼ $\frac{7.5 - y}{7.5} = \frac{be - de}{be}$

➼ $\frac{7.5 - y}{7.5} = \frac{bd}{be} ...... \: (1)$

Similarly ∆BDC ∼ ∆BEF

➼ $\frac{bd}{be} = \frac{cd}{ef} = \frac{bc}{bc + cf}$

➼ $\frac{bd}{be} = \frac{y}{4.5} = \frac{x}{x + 3} \: ....... \: (2)$

$From ( 1 ) and ( 2 )$

➼ $\frac{7.5 - y}{7.5} = \frac{y}{4.5}$

➼ $7.5y=(7.5)(4.5)-4.5y$

➼ $12y=(7.5)(4.5)=33.75$

➼ $y = \frac{33.75}{12} = 2.8125$

$From ( 2 )$

➼ $\frac{y}{4.5} = \frac{x}{x + 3}$

➼ $\frac{2.8}{4.5} = \frac{x}{x + 3}$

➼ $4.5 x = (2.8) x + 8.4$

➼$4.5 x - 2.8 x = 8.4$

➼ $1.7 x = 8.4$

➼ $x=\frac{8.4}{1.7}$

$\huge\pink{5\:(appr)}$

Answer 2

Answer:

• In △BCD and △BEF :

↠ ∠DBC = ∠FBE (Common angle)

↠ ∠BCD = ∠BEF (Corresponding angles of parallel lines CD and EF)

↠ ∠BDC = ∠BFE (Corresponding angles of parallel lines CD and EF)

∴ △BCD ∼△BEF (AAA rule)

Hence,

⇒ BE/BE = BF/BD = EF/CD (Corresponding sides)

⇒ x+3/x = BF/BD = 4.5/y ..(I)

Similarly, △FCD ∼△FAB

∴ y = 16/45 cm and, x = 5 cm