Question

What is the radius of the smallest circle?

Answer 1

Consider only the centers of circles

Let $AB=x=EQBC=yER$

$PD=AB+BC=x+y$

$(PD{)}^2+(PF{)}^2=(FD{)}^2$ (pythagoream theorem )

$FD=225+100=325$

$PF=225-100=125$

$P{D}^2=F{D}^2-P{F}^2={325}^2-{125}^2=\left(450\right)\times \left(200\right)$

$=900\times 100$

$PD^2=(300)^2

\ \Rightarrow \ PD=300=x+y$

$Q{E}^2+Q{F}^2=F{E}^2$

$\Rightarrow x^2=(225+r{)}^2-(225-r{)}^2$

$=(225+r+25-r)(225+r-225+r)$

$2\times 225\times 2\times r$

$x^e=30\sqrt{r}$

$(ER{)}^2+(RD{)}^2=F{D}^2$

$y^2=(100+r{)}^2-(100-r{)}^2$

$y=20\sqrt{r}$

$(x+y)=300$

$(x+y{)}^2=90000$

$x^2+y^2+2xy=90000\Rightarrow 900r+400r+1200r=90000$

$\Rightarrow 2500r=90000$

$\Rightarrow r=900/25\Rightarrow 36$