Question-225262

Question Number 225262 by fantastic last updated on 19/Oct/25

Commented by fantastic last updated on 19/Oct/25

F is midpoint of AB AG=GH=3 HB=2 Area of smaller circle?

$${F}\:{is}\:{midpoint}\:{of}\:{AB} \\ $$$${AG}={GH}=\mathrm{3} \\ $$$${HB}=\mathrm{2} \\ $$$${Area}\:{of}\:{smaller}\:{circle}? \\ $$

Commented by ajfour last updated on 20/Oct/25

So CD is diameter, i did fail to notice before. How's this video. Criticise somewhat, wanna better future ones. https://youtu.be/kNE2MWk-PlQ?si=cu3VdCj4sZwFZiiY

Commented by fantastic last updated on 20/Oct/25

$${sir}\:{if}\:{the}\:{CD}\:{length}\:{i} \\ $$$${think}\:{it}\:{will}\:{not}\:{pass} \\ $$$${G}\:{and}\:{H} \\ $$

Commented by fantastic last updated on 20/Oct/25

Answered by mr W last updated on 20/Oct/25

Commented by mr W last updated on 20/Oct/25

i didn′t say that it is the only path. certainly there are other methods. e.g. we can also get p^2 −0.5^2 =r^2 −1.5^2 ⇒4^2 −r^2 −0.5^2 =r^2 −1.5^2 ⇒r^2 =9

$${i}\:{didn}'{t}\:{say}\:\:{that}\:{it}\:{is}\:{the}\:{only}\:{path}. \\ $$$${certainly}\:{there}\:{are}\:{other}\:{methods}. \\ $$$${e}.{g}.\:{we}\:{can}\:{also}\:{get} \\ $$$${p}^{\mathrm{2}} −\mathrm{0}.\mathrm{5}^{\mathrm{2}} ={r}^{\mathrm{2}} −\mathrm{1}.\mathrm{5}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{4}^{\mathrm{2}} −{r}^{\mathrm{2}} −\mathrm{0}.\mathrm{5}^{\mathrm{2}} ={r}^{\mathrm{2}} −\mathrm{1}.\mathrm{5}^{\mathrm{2}} \:\Rightarrow{r}^{\mathrm{2}} =\mathrm{9} \\ $$

Commented by fantastic last updated on 20/Oct/25

$${thanks}\:{sir}. \\ $$

Commented by fantastic last updated on 20/Oct/25

$${this}\:{can}\:{be}\:{solved}\:{by} \\ $$$${only}\:{usimg}\:{Pythagoras} \\ $$$${Theorem} \\ $$

Commented by mr W last updated on 20/Oct/25

p^2 =4^2 −r^2 cos ∠GFE=((1^2 +p^2 −r^2 )/(2×1×p))=−((2^2 +p^2 −r^2 )/(2×2×p)) 2+p^2 −r^2 =0 2+4^2 −r^2 −r^2 =0 ⇒r^2 =9 ⇒r=3 area of small circle =πr^2 =9π ✓

$${p}^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$$\mathrm{cos}\:\angle{GFE}=\frac{\mathrm{1}^{\mathrm{2}} +{p}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}×\mathrm{1}×{p}}=−\frac{\mathrm{2}^{\mathrm{2}} +{p}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}×{p}} \\ $$$$\mathrm{2}+{p}^{\mathrm{2}} −{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{2}+\mathrm{4}^{\mathrm{2}} −{r}^{\mathrm{2}} −{r}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{r}^{\mathrm{2}} =\mathrm{9}\:\Rightarrow{r}=\mathrm{3} \\ $$$${area}\:{of}\:{small}\:{circle}\:=\pi{r}^{\mathrm{2}} =\mathrm{9}\pi\:\checkmark \\ $$

Commented by fantastic last updated on 20/Oct/25