Multiple choice

The area of the minor segment of a circle of radius 14 cm, when the angle of the corresponding sector is $60^0$, is

  1. $\frac{308}{3}-98\sqrt{3} cm^2$
  2. $\frac{308}{3}+49\sqrt{3} cm^2$
  3. $\frac{308}{3}+98\sqrt{3} cm^2$
  4. $\frac{308}{3}-49\sqrt{3} cm^2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Area of segment = Area of sector - Area of triangle. Area of sector = (60/360) * pi * 14^2 = (1/6) * (22/7) * 196 = (1/6) * 22 * 28 = 308/3. Area of triangle = (1/2) * r^2 * sin(60) = (1/2) * 196 * (sqrt(3)/2) = 49 * sqrt(3). Segment area = 308/3 - 49 * sqrt(3).

AI explanation

The area of a minor segment equals the area of the sector minus the area of the triangle, using pi times radius squared times theta divided by 360 for the sector. The sector area is 22 divided by 7 times 14 times 14 times 60 divided by 360, resulting in 308 divided by 3. The area of the equilateral triangle is root 3 divided by 4 times 14 times 14, which equals 49 times the square root of 3, so the segment area is 308 divided by 3 minus 49 times the square root of 3 cm squared.

Keep practicing — related questions