Answer :

**Given: **In the adjoining figure, ABCD is a trapezium in which CD || AB and its diagonals intersect at O. If AO = (5x – 7) cm, OC = (2x + 1) cm, DO = (7x – 5) cm and OB = (7x + 1) cm.**To find: **the value of x.**Solution:**

In the trapezium ABCD, AB || DC and its diagonals intersect at O.

Through O draw EO || AB meeting AD at E.

Now In Δ ADC

As EO || AB || DC

By thales theorem which states that If a line is drawn parallel to one side of a triangle to intersect the other

two sides in distinct points then the other two sides are divided in the same ratio.

...... (i)

In Δ DAB,

EO || AB

By thales theorem,

...... (ii)

From (i) and (ii)

⇒

⇒ (5x – 7)(7x + 1) = (7x – 5)(2x + 1)

⇒ 35x^{2} + 5x – 49x – 7 = 14x^{2} – 10x + 7x – 5

⇒ 35x^{2} – 44x – 7 = 14x^{2} – 3x – 5

⇒ 35x^{2} – 14x^{2} – 44x + 3x – 7 + 5 = 0

⇒ 21x^{2} – 41x – 2 = 0

⇒ 21x^{2} – 42x + x – 2 = 0

⇒ 21x(x – 2) + (x – 2) = 0

⇒ (21x + 1)(x – 2) = 0

⇒ (21x + 1) = 0 or (x – 2) = 0

⇒ x = -1/21 or x = 2

But x = -1/21 doesn’t satisfy the length of intersected lines.

So x ≠ -1/21

And thus, x = 2.

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