Question-228047

Question Number 228047 by mr W last updated on 14/Mar/26

Commented by mr W last updated on 14/Mar/26

$${find}\:{the}\:{area}\:{of}\:{the}\:{red}\:{circle}. \\ $$

Answered by TonyCWX last updated on 14/Mar/26

r = ((√(3^2 +3(3)+2(3)+2(3)(2)))/2) = (6/2) = 3 Area of circle = π(3)^2 = 9π

$${r}\:=\:\frac{\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{3}\left(\mathrm{3}\right)+\mathrm{2}\left(\mathrm{3}\right)+\mathrm{2}\left(\mathrm{3}\right)\left(\mathrm{2}\right)}}{\mathrm{2}}\:=\:\frac{\mathrm{6}}{\mathrm{2}}\:=\:\mathrm{3} \\ $$$$\mathrm{Area}\:\mathrm{of}\:\mathrm{circle}\:=\:\pi\left(\mathrm{3}\right)^{\mathrm{2}} \:=\:\mathrm{9}\pi \\ $$

Commented by TonyCWX last updated on 14/Mar/26

Commented by TonyCWX last updated on 14/Mar/26

$$\uparrow\:\mathrm{Generalisation}\:\mathrm{above} \\ $$

Answered by fantastic2 last updated on 14/Mar/26

Commented by fantastic2 last updated on 14/Mar/26

line_(blue) ^2 =0.5^2 +line_(green) ^2 4^2 −r^2 =0.5^2 +r^2 −1.5^2 16+2.25−0.25=2r^2 r^2 =9 area=πr^2 =9π

$${line}_{{blue}} ^{\mathrm{2}} =\mathrm{0}.\mathrm{5}^{\mathrm{2}} +{line}_{{green}} ^{\mathrm{2}} \\ $$$$\mathrm{4}^{\mathrm{2}} −{r}^{\mathrm{2}} =\mathrm{0}.\mathrm{5}^{\mathrm{2}} +{r}^{\mathrm{2}} −\mathrm{1}.\mathrm{5}^{\mathrm{2}} \\ $$$$\mathrm{16}+\mathrm{2}.\mathrm{25}−\mathrm{0}.\mathrm{25}=\mathrm{2}{r}^{\mathrm{2}} \\ $$$${r}^{\mathrm{2}} =\mathrm{9} \\ $$$${area}=\pi{r}^{\mathrm{2}} =\mathrm{9}\pi \\ $$

Answered by fantastic2 last updated on 14/Mar/26

Commented by fantastic2 last updated on 14/Mar/26

R=(1/2)(a+b+c) length of black line=(√(R^2 −r^2 )) length of red line=(√(r^2 −((b/2))^2 )) length of base of right triangle =(b/2)−R+a=(b/2)−((a+b+c)/2)+a=((a−c)/2) R^2 −r^2 =r^2 −((b/2))^2 +(((a−c)/2))^2 ⇒(((a+b+c)/2))^2 −(((a−c)/2))^2 +((b/2))^2 =2r^2 ⇒(1/4)(a^2 +b^2 +c^2 +2ab+2bc+2ca−a^2 −c^2 +2ac+b^2 )=2r^2 ⇒(1/4)(b^2 +ab+bc+2ac)=r^2 b=a=3 c=2 r^2 =(1/4)(9+9+6+12)=9

$${R}=\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{b}+{c}\right) \\ $$$${length}\:{of}\:{black}\:{line}=\sqrt{{R}^{\mathrm{2}} −{r}^{\mathrm{2}} } \\ $$$${length}\:{of}\:{red}\:{line}=\sqrt{{r}^{\mathrm{2}} −\left(\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$${length}\:{of}\:{base}\:{of}\:{right}\:{triangle} \\ $$$$=\frac{{b}}{\mathrm{2}}−{R}+{a}=\frac{{b}}{\mathrm{2}}−\frac{{a}+{b}+{c}}{\mathrm{2}}+{a}=\frac{{a}−{c}}{\mathrm{2}} \\ $$$${R}^{\mathrm{2}} −{r}^{\mathrm{2}} ={r}^{\mathrm{2}} −\left(\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{a}−{c}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left(\frac{{a}+{b}+{c}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{{a}−{c}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{b}}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{2}{r}^{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2}{bc}+\mathrm{2}{ca}−{a}^{\mathrm{2}} −{c}^{\mathrm{2}} +\mathrm{2}{ac}+{b}^{\mathrm{2}} \right)=\mathrm{2}{r}^{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}\left({b}^{\mathrm{2}} +{ab}+{bc}+\mathrm{2}{ac}\right)={r}^{\mathrm{2}} \\ $$$${b}={a}=\mathrm{3}\:{c}=\mathrm{2} \\ $$$${r}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{9}+\mathrm{9}+\mathrm{6}+\mathrm{12}\right)=\mathrm{9} \\ $$

Answered by mr W last updated on 14/Mar/26

Commented by mr W last updated on 14/Mar/26

a^2 =4^2 −r^2 −((1^2 +a^2 −r^2 )/(2×1×a))=((2^2 +a^2 −r^2 )/(2×2×a)) ⇒r^2 =((2×1^2 +2^2 +3×4^2 )/6)=9 red circle area =πr^2 =9π

$${a}^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$$−\frac{\mathrm{1}^{\mathrm{2}} +{a}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}×\mathrm{1}×{a}}=\frac{\mathrm{2}^{\mathrm{2}} +{a}^{\mathrm{2}} −{r}^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}×{a}} \\ $$$$\Rightarrow{r}^{\mathrm{2}} =\frac{\mathrm{2}×\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}×\mathrm{4}^{\mathrm{2}} }{\mathrm{6}}=\mathrm{9} \\ $$$${red}\:{circle}\:{area}\:=\pi{r}^{\mathrm{2}} =\mathrm{9}\pi \\ $$

Commented by A5T last updated on 14/Mar/26

Applying Stewart′s theorem to △BCD ⇒1×2×3+3a^2 =r^2 ×1+2r^2 ⇒6+48−3r^2 =3r^2 ⇒ r^2 =9

$$\mathrm{Applying}\:\mathrm{Stewart}'\mathrm{s}\:\mathrm{theorem}\:\mathrm{to}\:\bigtriangleup\mathrm{BCD} \\ $$$$\Rightarrow\mathrm{1}×\mathrm{2}×\mathrm{3}+\mathrm{3a}^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} ×\mathrm{1}+\mathrm{2r}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{6}+\mathrm{48}−\mathrm{3r}^{\mathrm{2}} =\mathrm{3r}^{\mathrm{2}} \:\Rightarrow\:\mathrm{r}^{\mathrm{2}} =\mathrm{9} \\ $$

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