Plz someone

find the area of the shaded portion in the equilateral triangle with sides 6. show all work for full credit.
(hint: assume that the central point of each arc is its corresponding vertex.)

Well you need to figure out the area of the circle piece. think of it as a separate thing. the radius is 3, and the angle of the circle piece is 60 (equilateral triangles are like that). So take the equation for area of a circle and times it by the fraction 60/360. this is 3³ × pi × (60÷360). this equals 1.5pi. since theres 3 of these little circle segments, times that by 3 (3*1.5pi = 4.5pi). then figure out the area of the triangle (it's 15.59) and subtract the circle area total. (15.59- 4.5pi). I left it as pi because i dont know what you need to round it to. 
Hope that helps!

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12