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Question

In the adjoining figure, ABCD is a trapezium in which ABDC. The diagonals AC and BD intersect at O. Prove that AO/OC=BO/OD.
1496571_35358858663541318c22fdf226796142.jpg

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Solution

From the given figure, ABCD is a trapezium in which ABDC,
The diagonals AC and BD intersect at O.
So we have to prove that, AO/OC=BO/OD
Consider the AOB and COD,
AOB=COD … [vertically opposite angles]
OAB=OCD
Therefore, AOBCOD
So, OA/OC=OB/OD
Now by using above result we have to find the value of x if OA=3x19,OB=x4,OC=x3 and OD=4.
OA/OC=OB/OD
(3x19)/(x3)=(x4)/4
By cross multiplication we get,
(x3)(x4)=4(3x19)
X24x3x+12=12x76
X27x+1212x+76=0
X219x+88=0
X28x11x+88=0
X(x8)11(x8)=0
(x8)(x11)=0
Take x8=0
X=8
Then, x11=0
X=11
Therefore, the value of x is 8 and 11.

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