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Question Number 7544 by Tawakalitu. last updated on 03/Sep/16

Commented by sou1618 last updated on 03/Sep/16

x=OA OB=(√(x^2 +144)) ΔABO=(1/2)12x ΔAOB=(1/2)3(12+x+(√(x^2 +144))) so 12x=3(12+x+(√(x^2 +144))) (√(x^2 +144))=3x−12 x^2 +144=9x^2 −72x+144 x(72−8x)=0 x=9 AB:AO=CE:CB 12:9=CE:12 CE=16 BE=(√(16^2 +12^2 ))=4×5=20 ΔBCE=(1/2)12×16=96 ΔBCE=(1/2)r(12+16+20)=24r so 96=24r r=4

$${x}={OA} \\ $$$${OB}=\sqrt{{x}^{\mathrm{2}} +\mathrm{144}} \\ $$$$\Delta{ABO}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{12}{x} \\ $$$$\Delta{AOB}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{3}\left(\mathrm{12}+{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{144}}\right) \\ $$$${so} \\ $$$$\mathrm{12}{x}=\mathrm{3}\left(\mathrm{12}+{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{144}}\right) \\ $$$$\sqrt{{x}^{\mathrm{2}} +\mathrm{144}}=\mathrm{3}{x}−\mathrm{12} \\ $$$${x}^{\mathrm{2}} +\mathrm{144}=\mathrm{9}{x}^{\mathrm{2}} −\mathrm{72}{x}+\mathrm{144} \\ $$$${x}\left(\mathrm{72}−\mathrm{8}{x}\right)=\mathrm{0} \\ $$$${x}=\mathrm{9} \\ $$$${AB}:{AO}={CE}:{CB} \\ $$$$\mathrm{12}:\mathrm{9}={CE}:\mathrm{12} \\ $$$${CE}=\mathrm{16} \\ $$$${BE}=\sqrt{\mathrm{16}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }=\mathrm{4}×\mathrm{5}=\mathrm{20} \\ $$$$\Delta{BCE}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{12}×\mathrm{16}=\mathrm{96} \\ $$$$\Delta{BCE}=\frac{\mathrm{1}}{\mathrm{2}}{r}\left(\mathrm{12}+\mathrm{16}+\mathrm{20}\right)=\mathrm{24}{r} \\ $$$${so} \\ $$$$\mathrm{96}=\mathrm{24}{r} \\ $$$${r}=\mathrm{4} \\ $$

Commented by sou1618 last updated on 03/Sep/16

Commented by Tawakalitu. last updated on 03/Sep/16

Wow, i really appreciate sir. God bless you.

$${Wow},\:{i}\:{really}\:{appreciate}\:{sir}.\:{God}\:{bless}\:{you}. \\ $$

Commented by Rasheed Soomro last updated on 03/Sep/16

I didn′t understand how ΔAOB=(1/2)3(12+x+(√(x^2 +144)))?

$${I}\:{didn}'{t}\:{understand}\:{how}\:\Delta{AOB}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{3}\left(\mathrm{12}+{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{144}}\right)? \\ $$

Commented by sandy_suhendra last updated on 03/Sep/16

the smaller circle is called the incircle of △ AOB the radius of an incircle : r = (A/s) ⇒ A=rs where A is the area △AOB and s is semiperimeter of △AOB so s=(1/2)(AO+AB+BO)

$${the}\:{smaller}\:{circle}\:{is}\:{called}\:{the}\:{incircle}\:{of}\:\bigtriangleup\:{AOB} \\ $$$${the}\:{radius}\:{of}\:{an}\:{incircle}\::\:{r}\:=\:\frac{{A}}{{s}}\:\Rightarrow\:{A}={rs} \\ $$$${where}\:{A}\:{is}\:{the}\:{area}\:\bigtriangleup{AOB}\:{and}\:{s}\:{is}\:{semiperimeter}\:{of}\:\bigtriangleup{AOB} \\ $$$${so}\:\:{s}=\frac{\mathrm{1}}{\mathrm{2}}\left({AO}+{AB}+{BO}\right) \\ $$

Commented by Rasheed Soomro last updated on 03/Sep/16

THαnkS !

$$\mathbb{TH}\alpha{n}\Bbbk\mathbb{S}\:! \\ $$

Commented by sandy_suhendra last updated on 03/Sep/16

using your picture and your answer for AO=9 and the rule of similarity ⇒△AOB∼△BCE so r : AO = R : BC ⇒r=radius of smaller circle R=radius of bigger circle 3 : 9 = R : 12 ⇒ R=4

$${using}\:{your}\:{picture}\:{and}\:{your}\:{answer}\:{for}\:{AO}=\mathrm{9} \\ $$$${and}\:{the}\:{rule}\:{of}\:{similarity}\:\Rightarrow\bigtriangleup{AOB}\sim\bigtriangleup{BCE} \\ $$$${so}\:\:\:{r}\::\:{AO}\:=\:{R}\::\:{BC}\:\Rightarrow{r}={radius}\:{of}\:{smaller}\:{circle} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{R}={radius}\:{of}\:{bigger}\:{circle} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{3}\::\:\mathrm{9}\:=\:{R}\::\:\mathrm{12}\:\Rightarrow\:{R}=\mathrm{4} \\ $$

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