in fig.4.144 we have AB||CD||EF.if AB=6cm , CD=x cm, EF=10cm, BD=4cm and DE=y cm, calculate the values of x and y

Consider ΔADB and ΔEDF,

∠ADB = ∠EDF (Vertical angles)
∠ABD = ∠FED (Alternate angles)
∠BAD = ∠EFD (Alternate angles)

∴ ΔADB ~ ΔEDF (ASA criterion)

⇒ BD/DE = AB/FE
⇒ 4/y = 6/10
⇒ 6y = 40
⇒ y = 40/6 = 20/3
⇒ y = 6.67

Similarly, ΔADE ~ ΔCDE (AA criterion)
⇒ DE/BE = DC/BA

⇒ 6.67 / 10.67 = x / 6

⇒ x = 6 x (6.67 / 10.67)

⇒ x = 3.75

hence, the values of x and y are 3.75 and 6.67.
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thisis

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y{1/x+1/z}=1 1/x+1/z=1/y Hence proved
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Consider ΔADB and ΔEDF, ∠ADB = ∠EDF (Vertical angles) ∠ABD = ∠FED (Alternate angles) ∠BAD = ∠EFD (Alternate angles) ∴ ΔADB ~ ΔEDF (ASA criterion) ⇒ BD/DE = AB/FE ⇒ 4/y = 6/10 ⇒ 6y = 40 ⇒ y = 40/6 = 20/3 ⇒ y = 6.67 Similarly, ΔADE ~ ΔCDE (AA criterion) ⇒ DE/BE = DC/BA ⇒ 6.67 / 10.67 = x / 6 ⇒ x = 6 x (6.67 / 10.67) ⇒ x = 3.75
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X=3.7 and y=6.6
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Answer is not right
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Please find this answer

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