A rectangular dance stage is lit by two lights that light up circular regions of the stage. The circles have radii of the same length and each circle passes through the center of the other. The stage perfectly circumscribes the two circles. A spectator throws a bouquet of flowers onto the stage. Assume the bouquet has an equal chance of landing anywhere on the stage.
a. What is the probability that the flowers land on a lit part of the stage?
b. What is the probability that the flowers land on the part of the stage where the spotlights overlap?
+37170 Area of stage will be: STAGE AREA= 3r x 2r = 6r^2 From the description and diagram
The LIT area will be
pi r^2 + (pi r^2 - 2 x (segment of overlap) ) (1)
Segment of overlap is portended by an angle of 120 degrees
120/360 (pi r^2) = 1/3 pi r^2 is the area of the SECTOR portended by the 120 degrees
we will have to SUBTRACT TWO TIMES the area of a triangle with hypotenuse = r and 1 leg = 1/2R to find the area of the SEGMENT
Using Pythagorean theorem to find the other leg r^2 - (1/2r)^2 = (other leg)^2
Other leg = sqrt(r^2-(1/2r^)^2 ) = sqrt(r^2-1/4r^2) = sqrt(3/4r^2)= r (sqrt3)/2
So after all of this, the area of this RIGHT triangle is 1/2 (Leg1 x Leg2) = 1/2 ( 1/2 r x r (sqrt3)/2) = 1/2 ( r^2 (sqrt3)/4) = r^2 sqrt3 / 8
NOW,(hang in there.....this get's long) TWO times this triangle is ( r^2sqrt3) /4
Area of SECTOR - TWO TIMES TRIANGLE = area of SEGMENT = 1/3 pi r^2 -( r^2 sqrt 3 ) /4 (2)
Getting there: from (1) the LIT area of the stage is pi r^2 + pi r^2 - 2 (1/3 pir^2- (r^2sqrt3)/4)
2 pi r^2 - 2/3 pi r^2 - r^2 (sqrt3)/2 = 4/3 pi r^2 -r^2 (sqrt3)/2
Now the PERCENTAGE of the stage is LIT area/ stage area 4/3 pi r^2 -r^2 (sqrt3)/2 / 6r^2
cancel out all of those r^2 to get: ( 4/3 pi - (sqrt3)/2 ) / 6 = 55.4% of the stage is lit 55.4 percent change flowers will land in the light
b) we found the are of the SECTOR to be (above--- (2) ) 1/3 pi r^2 -( r^2 sqrt 3 ) /4 the are of over lap is TWO times this
so the are of stage in overlapping light is 2 x sectors/ stage area = 2(1/3 pi r^2 -( r^2 sqrt 3 ) /4) / 6r^2 cancel out all of th r^2 to get
(2/3pi -( sqrt3)/2) / 6 = 20.47 percent chance of landing in overlap area
Wow.....there was a LOT of calcs to get this answer......hope I didn't mess up in there somewhere ! Anyone else get same answers???
