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Question Number 166861 by Tawa11 last updated on 01/Mar/22

Commented by cortano1 last updated on 01/Mar/22

$$\Rightarrow 3.\mathrm{x}=6.2\Rightarrow \mathrm{x}=4$$$$\Rightarrow \mathrm{r}=\sqrt{2^2+6^2+3^2+4^2}=\sqrt{49+16}=\sqrt{65}$$

Commented by mr W last updated on 01/Mar/22

$$wrong.$$$$r=\frac{\sqrt{65}}{2}$$

Commented by cortano1 last updated on 01/Mar/22

$$\mathrm{AB}\mathrm{not}\mathrm{diameter}$$

Commented by cortano1 last updated on 01/Mar/22

Commented by mr W last updated on 01/Mar/22

$$i{didn}^{\prime }tsayABisdiameter,seebelow.$$

Commented by mr W last updated on 01/Mar/22

$$a^2+b^2+c^2+d^2=R^{2}iscorrect,butit$$$$isalsosaid\left(inJapanese\right)thatR$$$$isthediameterofthecircle.$$

Commented by Tawa11 last updated on 01/Mar/22

$$\mathrm{Thanks}\mathrm{sir}.\mathrm{I}\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Answered by TheSupreme last updated on 01/Mar/22

$$ADBrectangle$$$$AD=\sqrt{4+9}$$$$DB=\sqrt{9+36}$$$$AB=\sqrt{4+9+9+36}=\sqrt{58}$$$$r=\frac{\sqrt{58}}{2}$$

Commented by mr W last updated on 01/Mar/22

$$wrong.$$$$howareyousurethat\mathrm{\angle }ADBis$$$$rectangleandABisdiameter?$$$$ifABwerediameterand\mathrm{\angle }ADB$$$$wererectangle,thenyoushould$$$$have2\times 6=3^2,butthisisnotsure,$$$$thereforeABisnotdiameterand$$$$\mathrm{\angle }ADBisnotrectangle.$$

Answered by mr W last updated on 01/Mar/22

Commented by mr W last updated on 01/Mar/22

$$AE=EB=\frac{2+6}{2}=4$$$$FG=OE=\sqrt{r^2-4^2}$$$$OF=EG=AE-AG=4-2=2$$$$FD=DG+FG=3+\sqrt{r^2-4^2}$$$${OD}^2={FD}^2+{OF}^2$$$$r^2={(3+\sqrt{r^2-4^2})}^2+2^2$$$$9+6\sqrt{r^2-4^2}-4^2+2^2=0$$$$\sqrt{r^2-4^2}=\frac{1}{2}$$$$r=\sqrt{4^2+{\left(\frac{1}{2}\right)}^2}=\frac{\sqrt{65}}{2}\approx 4.031$$

Commented by Tawa11 last updated on 01/Mar/22

$$\mathrm{Thanks}\mathrm{sir},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Answered by MJS_new last updated on 04/Mar/22

$$\mathrm{the}\mathrm{circle}\mathrm{is}\mathrm{the}\mathrm{circumcircle}\mathrm{of}\mathrm{the}△ABD$$$$\mathrm{with}\mathrm{sides}\mid AB\mid =8;\mid BD\mid =3\sqrt{5};\mid AD\mid =\sqrt{13}$$$$\Rightarrow$$$$r=\frac{abc}{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}=\frac{\sqrt{65}}{2}$$

Commented by Tawa11 last updated on 01/Mar/22

$$\mathrm{Thanks}\mathrm{sir},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by MJS_new last updated on 04/Mar/22

$$\mathrm{yes},\mathrm{thank}\mathrm{you}!$$

Commented by malwan last updated on 04/Mar/22

$$greatsir$$$$butIthinkthethirdbracket$$$$is(a-b+c)$$

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