In the given figure DEI BC, find the value of x
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Answered by
6
DE is parallel to BC in Triangle ABC
So by Thales theorem
AD/BD = AE/EC
X/3X+4 = x+3/3x+19
Cross multiplying ,
3x^2+19x = 3x^2+13x+12
6x=12
X=2
So by Thales theorem
AD/BD = AE/EC
X/3X+4 = x+3/3x+19
Cross multiplying ,
3x^2+19x = 3x^2+13x+12
6x=12
X=2
Answered by
1
Answer:
value of X is 2
Step-by-step explanation:
By BPT theorem,
AD/BD=AE/CE
x/3x+4=X+3/3x+19
By solving this equation,
you get the and as 2 which is the correct value of X as required.
Hope this helps you
Thanks
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