Multiple choice

Two parallel chords $AB$ and $CD$ are drawn on the same side of centre $O$ of a circle. Radius of circle is $65: m$, length of chords are $112: m$ and $126: m$. Find the area of $\square ABCD$.

  1. $2023\:m^2$
  2. $2123\:m^2$
  3. $2223\:m^2$
  4. $2323\:m^2$
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A Correct answer
Explanation

The distance of the chords from the center are found using the Pythagorean theorem: d1 = sqrt(65^2 - 56^2) = 33 and d2 = sqrt(65^2 - 63^2) = 16. Since they are on the same side, the distance between them is 33 - 16 = 17. The area of the trapezoid ABCD is ((AB + CD) / 2) * height = ((112 + 126) / 2) * 17 = 119 * 17 = 2023.

AI explanation

For a circle of radius 65 m, a chord of length 112 m is located at a distance of sqrt(65^2 - 56^2) = 33 m from the center. A chord of 126 m is located at a distance of sqrt(65^2 - 63^2) = 16 m from the center. Since the chords are on the same side of the center, the distance between them is 33 - 16 = 17 m. The area of trapezium ABCD is found by multiplying the height by the average of the parallel sides, which is 17 * (112 + 126) / 2 = 2023 m^2. The area of the quadrilateral is 2023 m^2.

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