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

Commented by sou1618 last updated on 03/Sep/16

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

Commented by sou1618 last updated on 03/Sep/16

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

$$Wow,ireallyappreciatesir.Godblessyou.$$

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

$$I{didn}^{\prime }tunderstandhow\mathrm{\Delta }AOB=\frac{1}{2}3(12+x+\sqrt{x^2+144})?$$

Commented by sandy_suhendra last updated on 03/Sep/16

$$thesmallercircleiscalledtheincircleof△AOB$$$$theradiusofanincircle:r=\frac{A}{s}\Rightarrow A=rs$$$$whereAisthearea△AOBandsissemiperimeterof△AOB$$$$sos=\frac{1}{2}(AO+AB+BO)$$

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

$$\mathbb{TH}\alpha n\mathbb{k}\mathbb{S}!$$

Commented by sandy_suhendra last updated on 03/Sep/16

$$usingyourpictureandyouranswerforAO=9$$$$andtheruleofsimilarity\Rightarrow △AOB\sim △BCE$$$$sor:AO=R:BC\Rightarrow r=radiusofsmallercircle$$$$R=radiusofbiggercircle$$$$3:9=R:12\Rightarrow R=4$$

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