Question 773501
I can only draw full circle. I cannot draw semicircles, so pretend you do not see the wromg half of the circles.
{{{drawing(300,300,-2,5,-2.5,4.5,
circle(2,1.5,2.5),
circle(0,1.5,1.5),
circle(2,0,2),
red(triangle(0,0,0,3,4,0)),
red(rectangle(0,0,0.2,0.2)),
locate(-0.2,0,P),locate(4,0,Q),
locate(-0.2,3.3,R),locate(2,0,r),
locate(0.05,1.8,q),locate(2,1.5,p)
)}}} According to Pythagoras, {{{r^2+q^2=p^2}}}.
The area of a circle is
{{{area=pi*radius^2}}}
so the area of a smicircle is {{{area=(1/2)*pi*radius^2}}}
The circles' diameters are {{{r}}}, {{{q}}} and {{{p}}}, so their radii are {{{r/2}}}, {{{q/2}}} and {{{p/2}}}.
Then the areas of the semicircles are
{{{(1/2)*pi*(r/2)^2}}}, {{{(1/2)*pi*(q/2)^2}}} and {{{(1/2)*pi*(p/2)^2}}}.
The sum of the areas of the semicircles is
{{{(1/2)*pi*(r/2)^2+(1/2)*pi*(q/2)^2+(1/2)*pi*(p/2)^2=(1/2)*pi*((r/2)^2+(q/2)^2+(p/2)^2)=(1/2)*pi*(r^2/4+q^2/4+p^2/4)=(1/2)*pi*((r^2+q^2+p^2)/4)}}}
Substituting {{{p^2}}} for {{{r^2+q^2}}}, we get
sum of semicircle areas = {{{(1/2)*pi*((p^2+p^2)/4)=(1/2)*pi*(2p^2/4)=(1/2)*pi*(p^2/2)=highlight((1/4)*pi*p^2)}}}