Please solve question 25
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JenlliaMishra:
By the way in which school u r reading?
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X =2cm and y=6cm as I think
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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
hence, the values of x and y are 3.75 and 6.67.
HOPE IT HELPS U IF YEAH PLZ MARK IT AS BRAINLIEST❤️✌️
∠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.
HOPE IT HELPS U IF YEAH PLZ MARK IT AS BRAINLIEST❤️✌️
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