AB is a chord with centre O and radius 4 cm. AB is of length 4 cm and divides the circle into two segments. Find d area of minor segment.
A chord PQ of length 12cm suntends an angle of 120 degree at the centre of a circle. Find the area of minor segment cut off by the chord PQ
To find the area of the minor segment, we need to determine the area of the sector formed by the chord and subtract the area of the triangle formed by the chord.
1. Start by finding the area of the sector:
- The radius of the circle is 4 cm, which means the length of AO (or OB) is also 4 cm.
- The central angle AOB is 360 degrees since it is a complete circle.
- The arc length AB is equal to the length of the chord, which is 4 cm.
- To find the sector area, we can use the formula: Area of sector = (central angle / 360) * π * radius^2.
- For the given chord AB, the central angle is 120 degrees.
- So, the area of the sector = (120/360) * π * 4^2 = (1/3) * π * 16 = (16/3)π cm^2.
2. Next, find the area of the triangle:
- The triangle formed by the chord AB is an equilateral triangle because both sides are of length 4 cm, and AB is a chord.
- The formula to calculate the area of an equilateral triangle is: Area = (sqrt(3) / 4) * side^2.
- In this case, the side length is 4 cm.
- So, the area of the triangle = (sqrt(3) / 4) * 4^2 = (sqrt(3) / 4) * 16 = 4sqrt(3) cm^2.
3. Finally, subtract the area of the triangle from the area of the sector to find the area of the minor segment:
- Area of minor segment = Area of sector - Area of triangle
- Area of minor segment = (16/3)π - 4sqrt(3) cm^2.
Therefore, the area of the minor segment is (16/3)π - 4sqrt(3) cm^2.
area of segment is
a = 1/2 r^2 (θ-sinθ)
here,
4 sin θ/2 = 2, so θ=π/3
so,
a = 1/2 (16) (π/3 - √3/2)
= 4/3 (2π-3√3)
use same formula with θ=2π/3, r*sin(θ/2)=6 to find r