5. In the given figure ABCD is a trapezium with AB||CD. If AO = x – 1, CO = BO = x + 1 and OD = x + 4, find the value of x.
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From given statement:
In △ADC
EO∣∣AB∣∣DC
By thales theorem:
ED
AE
=
OC
AO
...(1)
In △DAB
EO∣∣AB
So, By thales theorem:
EA
DE
=
OB
DO
...(2)
From (1) and (2)
OC
AO
=
OB
DO
(2x+1)
(5x−7)
=
(7x+1)
(7x−5)
(5x−7)(7x+1)=(7x−5)(2x+1)
35x
2
+5x–49x–7=14x
2
–10x+7x−5
35x
2
−14x
2
–44x+3x−7+5=0
21x
2
−42x+x−2=0
21(x−2)+(x−2)=0
(21x+1)(x−2)=0
Either (21x+1)=0 or (x–2)=0
x=
21
−1
(does not satisfy) or x=2
⇒x=2.
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