Multiple choice

A circle of radius 6 cm is placed overlapping a square of area 128 cm2, such that the centre of the circle lies at the point of intersection of the diagonals of the square. What is the maximum distance between any vertex of the square and any point on the circumference of the circle?

  1. 10 cm

  2. 12 cm

  3. 14 cm

  4. 16 cm

  5. 18 cm
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Area of square = 128, so side length s = sqrt(128) = 8 * sqrt(2). Distance from center to vertex = (1/2) * diagonal = (1/2) * (s * sqrt(2)) = (1/2) * (8 * sqrt(2) * sqrt(2)) = 8 cm. The circle has radius 6 cm. The maximum distance from a vertex to any point on the circle is the distance from the vertex to the center plus the radius: 8 + 6 = 14 cm.

AI explanation

The side length of the square is √128 = 8√2 cm. The distance from the center to a vertex is half the diagonal, calculated as (8√2 * √2) / 2 = 8 cm. The maximum distance from the vertex to the circumference is the distance to the center plus the radius, which is 8 + 6 = 14 cm.

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