Question
Download Solution PDFAB is parallel to DC in a trapezium ABCD. It is given that AB > DC and the diagonals AC and BD intersect at O. If AO = 3x − 15, OB = x + 9, OC
= x - 5 and OD = 5, and x has two values x1 and x2, then the value of \(\left(x_1^2+x_2^2\right)\) is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
AB is parallel to DC in a trapezium ABCD. AB > DC and the diagonals AC and BD intersect at O.
AO = 3x - 15, OB = x + 9, OC = x - 5, OD = 5
Formula used:
In a trapezium with parallel sides AB and DC, the diagonals intersecting at O divide each other proportionally:
\(\dfrac{AO}{OB} = \dfrac{OC}{OD}\)
Calculation:
\(\dfrac{3x-15}{x+9} = \dfrac{x-5}{5}\)
⇒ (3x - 15) × 5 = (x + 9) × (x - 5)
⇒ 15x - 75 = x2 + 9x - 5x - 45
⇒ 15x - 75 = x2 + 4x - 45
⇒ x2 - 11x + 30 = 0
Solving the quadratic equation:
⇒ x =\(\frac{11 \pm \sqrt{121 - 120}}{2}\)
⇒ x1 = 6, x2 = 5
To find x12 + x22:
⇒ 62 + 52 = 36 + 25 = 61
∴ The correct answer is option (3).
Last updated on May 28, 2025
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