Multiple choice

If a chord $AB$ of length $8\ cm$ is making $90^{o}$ at the centre $'O'$ of a circle, the find the area of $\triangle AOB$ (in $cm^{2}$)

  1. $10$
  2. $14$
  3. $16$
  4. $18$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

OA and OB are radii, and angle AOB is 90 degrees, so triangle AOB is a right isosceles triangle. Since AB = 8, its area is half the product of the perpendicular radii, which equals 16 square centimetres.

AI explanation

Since chord AB makes a 90 degree angle at center O, triangle AOB is a right-angled isosceles triangle. Let the radius be r, so OA = OB = r. Using the Pythagorean theorem gives r^2 + r^2 = 8^2, meaning 2r^2 = 64 and r^2 = 32. The area of triangle AOB is (1/2) * base * height, which equals (1/2) * r * r = (1/2) * r^2 = (1/2) * 32 = 16 cm^2.

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