in figure given below De parallel to BC find the value of x..pls guys give answer...no spam right answer I want
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Given : A Triangle ABC in which DE ║ BC AD = x , BD = x + 1 , AC = x + 3, CE = x + 5
To find : Value of x
Solution:
DE ║ BC
=> AD/BD = AE/CE
=> x/(x +1 ) = (x + 3)/(x + 5)
=> x(x + 5) = (x + 3)(x + 1)
=> x² + 5x = x² + 4x + 3
=> x = 3
Value of x = 3
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Answer:
DE II BC
Angle E =Angle B= 50° (Transversal)
Angle ADE =180°-95°=85°
Therefore, Angle x+50°+85°=180°
Angle x=180°-135°=45°
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