Question-205861

Question Number 205861 by mr W last updated on 01/Apr/24

Answered by A5T last updated on 01/Apr/24

9×9=(2r−x)x⇒2r−x=((81)/x) 21×21=(2r−16−x)(16+x)=(((81)/x)−16)(16+x) ⇒x=2⇒r=21.25

$$\mathrm{9}×\mathrm{9}=\left(\mathrm{2}{r}−{x}\right){x}\Rightarrow\mathrm{2}{r}−{x}=\frac{\mathrm{81}}{{x}} \\ $$$$\mathrm{21}×\mathrm{21}=\left(\mathrm{2}{r}−\mathrm{16}−{x}\right)\left(\mathrm{16}+{x}\right)=\left(\frac{\mathrm{81}}{{x}}−\mathrm{16}\right)\left(\mathrm{16}+{x}\right) \\ $$$$\Rightarrow{x}=\mathrm{2}\Rightarrow{r}=\mathrm{21}.\mathrm{25} \\ $$

Commented by mr W last updated on 01/Apr/24

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Answered by mr W last updated on 01/Apr/24

(√(R^2 −(12+9)^2 ))+16=(√(R^2 −9^2 )) R^2 −21^2 +16^2 +32(√(R^2 −21^2 ))=R^2 −9^2 4(√(R^2 −21^2 ))=13 R=(√(21^2 +(((13)/4))^2 ))=((85)/4)=21.25

$$\sqrt{{R}^{\mathrm{2}} −\left(\mathrm{12}+\mathrm{9}\right)^{\mathrm{2}} }+\mathrm{16}=\sqrt{{R}^{\mathrm{2}} −\mathrm{9}^{\mathrm{2}} } \\ $$$${R}^{\mathrm{2}} −\mathrm{21}^{\mathrm{2}} +\mathrm{16}^{\mathrm{2}} +\mathrm{32}\sqrt{{R}^{\mathrm{2}} −\mathrm{21}^{\mathrm{2}} }={R}^{\mathrm{2}} −\mathrm{9}^{\mathrm{2}} \\ $$$$\mathrm{4}\sqrt{{R}^{\mathrm{2}} −\mathrm{21}^{\mathrm{2}} }=\mathrm{13} \\ $$$${R}=\sqrt{\mathrm{21}^{\mathrm{2}} +\left(\frac{\mathrm{13}}{\mathrm{4}}\right)^{\mathrm{2}} }=\frac{\mathrm{85}}{\mathrm{4}}=\mathrm{21}.\mathrm{25} \\ $$

Commented by Skabetix last updated on 02/Apr/24

$$ \\ $$Hello sir, how did you arrive at the first equation? Which theorems did you use? Congratulations anyway.

Commented by mr W last updated on 02/Apr/24

Commented by mr W last updated on 02/Apr/24

AB=(√(R^2 −(12+9)^2 )) AC=(√(R^2 −9^2 )) AB+16=AC (√(R^2 −(12+9)^2 ))+16=(√(R^2 −9^2 ))

$${AB}=\sqrt{{R}^{\mathrm{2}} −\left(\mathrm{12}+\mathrm{9}\right)^{\mathrm{2}} } \\ $$$${AC}=\sqrt{{R}^{\mathrm{2}} −\mathrm{9}^{\mathrm{2}} } \\ $$$${AB}+\mathrm{16}={AC} \\ $$$$\sqrt{{R}^{\mathrm{2}} −\left(\mathrm{12}+\mathrm{9}\right)^{\mathrm{2}} }+\mathrm{16}=\sqrt{{R}^{\mathrm{2}} −\mathrm{9}^{\mathrm{2}} } \\ $$

Answered by mr W last updated on 01/Apr/24

Commented by mr W last updated on 01/Apr/24

R=((AC)/(2 sin (π−θ))) =((√(16^2 +(12+9+9)^2 ))/(2×((16)/( (√(16^2 +12^2 ))))))=((85)/4)=21.15

$${R}=\frac{{AC}}{\mathrm{2}\:\mathrm{sin}\:\left(\pi−\theta\right)} \\ $$$$\:\:\:\:=\frac{\sqrt{\mathrm{16}^{\mathrm{2}} +\left(\mathrm{12}+\mathrm{9}+\mathrm{9}\right)^{\mathrm{2}} }}{\mathrm{2}×\frac{\mathrm{16}}{\:\sqrt{\mathrm{16}^{\mathrm{2}} +\mathrm{12}^{\mathrm{2}} }}}=\frac{\mathrm{85}}{\mathrm{4}}=\mathrm{21}.\mathrm{15} \\ $$

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