Answer:
1π
Explanation:
suppose the radius of semicircle P is r,
then the radius of semicircle Q = (r+d)/2 ... d≤r
radius of semicircle R = (r-d)/2
area P = 1/2 (r)²π
area Q = 1/2 ((r+d)/2)² π = 1/8 (r² + 2rd + d²)π
area R = 1/2 ((r-d)/2)² π = 1/8 (r² - 2rd + d²)π
shaded area = P-Q-R = 1/2 r²π - 1/4 (r² + d²)π
= ((r² - d²)/4) * π
because there is no constant r value in the question and d value changes with the r change, when the vertical segment length equal the semicircle P radius (r), r=2 and d = 0
therefore the shaded area = ((2² -0²)/4)*π = 1π