Multiple choice

In a circle of diameter 50cm chords AB and CD are drawn parallel such that sum of their distance is 31cm and difference between their distances is 17cm.Then the length of each of the chords and A($\square$ABCD) in both the condition if AB < CD are respectively

  1. AB = 7cm, CD = 48cm, A($\square$ABCD) = 961c$m^2$ and 527 c$m^2$
  2. AB = 14cm, CD = 48cm, A($\square$ABCD) = 527c$m^2$ and 961 c$m^2$
  3. AB = 7cm, CD = 24cm, A($\square$ABCD) = 527/2c$m^2$ and 961/2 c$m^2$
  4. AB = 48cm, CD = 14cm, A($\square$ABCD) = 527c$m^2$ and 961 c$m^2$
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B Correct answer
Explanation

For a circle of radius 25 cm, a chord at distance d from the center has length 2sqrt(625 - d^2). The two distances are 24 cm and 7 cm, giving chord lengths 14 cm and 48 cm, respectively. Since AB < CD, AB = 14 cm and CD = 48 cm, with the corresponding trapezium areas 527 cm^2 and 961 cm^2.

AI explanation

The circle has a radius of 25 cm, and solving the given sum and difference equations yields distances from the center of 7 cm and 24 cm. Using the perpendicular bisector property of chords, the lengths are 2*sqrt(25^2 - 7^2) = 48 cm and 2*sqrt(25^2 - 24^2) = 14 cm. Since AB is the shorter chord, AB is 14 cm and CD is 48 cm. Connecting the endpoints of these parallel chords forms a trapezoid, where using the formula for the area of a trapezoid yields 527 cm^2 if they are on the same side of the center, and 961 cm^2 if they are on opposite sides.

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