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Home » In the adjoining figure, ABCD is a trapezium in which AB || DC. The diagonals AC and BD intersect at O. Prove that AO/OC = BO/ODUsing the above result, find the values of x if OA = 3x – 19, OB = x – 4, OC = x – 3 and OD = 4.
In the adjoining figure, ABCD is a trapezium in which AB || DC. The diagonals AC and BD intersect at O. Prove that AO/OC = BO/ODUsing the above result, find the values of x if OA = 3x – 19, OB = x – 4, OC = x – 3 and OD = 4.
CBSE | Class 10 | maths | Maths | ML Aggarwal | Similarity
In the adjoining figure, ABCD is a trapezium in which AB || DC. The diagonals AC and BD intersect at O. Prove that AO/OC = BO/ODUsing the above result, find the values of x if OA = 3x – 19, OB = x – 4, OC = x – 3 and OD = 4.

Solution:-

From the given figure, ABCD is a trapezium in which AB || DC,

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, ∆AOB ~ ∆COD

So, OA/OC = OB/OD

Now by using above result we have to find the value of x if OA = 3x – 19, OB = x – 4, OC = x – 3 and OD = 4.

OA/OC = OB/OD

(3x – 19)/(x – 3) = (x – 4)/4

By cross multiplication we get,

(x – 3) (x – 4) = 4(3x – 19)

X2 – 4x – 3x + 12 = 12x – 76

X2 – 7x + 12 – 12x + 76 = 0

X2 – 19x + 88 = 0

X2 – 8x – 11x + 88 = 0

X(x – 8) – 11(x – 8) = 0

(x – 8) (x – 11) = 0

Take x – 8 = 0

X = 8

Then, x – 11= 0

X = 11

Therefore, the value of x is 8 and 11.

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