PQ is parallel to SR in a trapezium PQRS. It is given that PQ > SR and the diagonals PR and QS intersect at O. If PO = 3x − 15, OQ = x + 9, OR = x − 5 and OS = 5, and x have two values x1 and x2, then the value of (x12 - x22) is:
Show answer & explanation
Correct answer: B
Explanation
The correct answer is option 2 i.e. 11.
We know that,
When diagonals PR and QS are intersecting at O, the ratio of the segments of the diagonals is equal.
PO/OR = OQ/OS
According to the question;
⇒ (3x − 15)/(x - 5) = (x + 9)/(5)
⇒ 15x - 75 = x2 + 4x - 45
⇒ x2 - 11x + 30 = 0
Hence, x2 = 5 and x1 = 6
Therefore, the required value = (x12 - x22) = 62 - 52 = 36 - 25 = 11