Question-221788

Question Number 221788 by Tawa11 last updated on 10/Jun/25

Answered by mr W last updated on 10/Jun/25

Commented by mr W last updated on 10/Jun/25

AC=(√((2R)^2 −6^2 )) BD=(√((2R)^2 −14^2 )) AC×BD=6×(2R)+14×6 (√((4R^2 −36)(4R^2 −196)))=12R+14×6 (√((R^2 −9)(R^2 −49)))=3(R+7) R^2 −7R−18=0 (R+2)(R−9)=0 ⇒R=9 ✓

$${AC}=\sqrt{\left(\mathrm{2}{R}\right)^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} } \\ $$$${BD}=\sqrt{\left(\mathrm{2}{R}\right)^{\mathrm{2}} −\mathrm{14}^{\mathrm{2}} } \\ $$$${AC}×{BD}=\mathrm{6}×\left(\mathrm{2}{R}\right)+\mathrm{14}×\mathrm{6} \\ $$$$\sqrt{\left(\mathrm{4}{R}^{\mathrm{2}} −\mathrm{36}\right)\left(\mathrm{4}{R}^{\mathrm{2}} −\mathrm{196}\right)}=\mathrm{12}{R}+\mathrm{14}×\mathrm{6} \\ $$$$\sqrt{\left({R}^{\mathrm{2}} −\mathrm{9}\right)\left({R}^{\mathrm{2}} −\mathrm{49}\right)}=\mathrm{3}\left({R}+\mathrm{7}\right) \\ $$$${R}^{\mathrm{2}} −\mathrm{7}{R}−\mathrm{18}=\mathrm{0} \\ $$$$\left({R}+\mathrm{2}\right)\left({R}−\mathrm{9}\right)=\mathrm{0}\:\Rightarrow{R}=\mathrm{9}\:\checkmark \\ $$

Commented by Tawa11 last updated on 10/Jun/25

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

Answered by mr W last updated on 10/Jun/25

Commented by mr W last updated on 10/Jun/25

R^2 −7^2 =6^2 −(R−7)^2 R^2 −7R−18=0 (R+2)(R−9)=0 ⇒R=9 ✓

$${R}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} =\mathrm{6}^{\mathrm{2}} −\left({R}−\mathrm{7}\right)^{\mathrm{2}} \\ $$$${R}^{\mathrm{2}} −\mathrm{7}{R}−\mathrm{18}=\mathrm{0} \\ $$$$\left({R}+\mathrm{2}\right)\left({R}−\mathrm{9}\right)=\mathrm{0}\:\Rightarrow{R}=\mathrm{9}\:\checkmark \\ $$

Commented by Tawa11 last updated on 10/Jun/25

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

Answered by mehdee7396 last updated on 10/Jun/25

((14)/(sin2θ))=(r/(sinθ)) & (6/(sinθ))=(r/(sin(90−(θ/2)))) (7/(cosθ))=r⇒(6/(sinθ))=(7/(cosθcos(θ/2))) 7sin(θ/2)=3cosθ⇒6sin^2 (θ/2)+7sin(θ/2)−3=0 ⇒sin(θ/2)=(1/3)⇒cosθ=(7/9) ⇒r=9

$$\frac{\mathrm{14}}{{sin}\mathrm{2}\theta}=\frac{{r}}{{sin}\theta}\:\:\:\&\:\:\:\frac{\mathrm{6}}{{sin}\theta}=\frac{{r}}{{sin}\left(\mathrm{90}−\frac{\theta}{\mathrm{2}}\right)} \\ $$$$\frac{\mathrm{7}}{{cos}\theta}={r}\Rightarrow\frac{\mathrm{6}}{{sin}\theta}=\frac{\mathrm{7}}{{cos}\theta{cos}\frac{\theta}{\mathrm{2}}} \\ $$$$\mathrm{7}{sin}\frac{\theta}{\mathrm{2}}=\mathrm{3}{cos}\theta\Rightarrow\mathrm{6}{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}+\mathrm{7}{sin}\frac{\theta}{\mathrm{2}}−\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow{sin}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow{cos}\theta=\frac{\mathrm{7}}{\mathrm{9}}\:\Rightarrow{r}=\mathrm{9}\:\: \\ $$$$ \\ $$

Commented by mehdee7396 last updated on 10/Jun/25

Commented by Tawa11 last updated on 10/Jun/25

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

Commented by Tawa11 last updated on 10/Jun/25

$$\mathrm{How}\:\mathrm{do}\:\mathrm{you}\:\mathrm{arrive}\:\mathrm{at}\:\:\mathrm{7sin}\frac{\theta}{\mathrm{2}}\:\:=\:\:\mathrm{3cos}\theta \\ $$

Commented by mehdee7396 last updated on 10/Jun/25

(6/(sinθ))=(7/(cosθcos(θ/2))) ⇒(6/(2sin(θ/2)cos(θ/2)))=(7/(cosθcos(θ/2))) ⇒6cosθcos(θ/2)=14sin(θ/2)cos(θ/2) ⇒3cosθ=7sin(θ/2)

$$\frac{\mathrm{6}}{{sin}\theta}=\frac{\mathrm{7}}{{cos}\theta{cos}\frac{\theta}{\mathrm{2}}} \\ $$$$\Rightarrow\frac{\mathrm{6}}{\mathrm{2}{sin}\frac{\theta}{\mathrm{2}}{cos}\frac{\theta}{\mathrm{2}}}=\frac{\mathrm{7}}{{cos}\theta{cos}\frac{\theta}{\mathrm{2}}} \\ $$$$\Rightarrow\mathrm{6}{cos}\theta{cos}\frac{\theta}{\mathrm{2}}=\mathrm{14}{sin}\frac{\theta}{\mathrm{2}}{cos}\frac{\theta}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{3}{cos}\theta=\mathrm{7}{sin}\frac{\theta}{\mathrm{2}} \\ $$$$ \\ $$

Commented by Tawa11 last updated on 10/Jun/25

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{understand}\:\mathrm{now}. \\ $$

Answered by mr W last updated on 10/Jun/25

Commented by mr W last updated on 11/Jun/25

$${yes}.\:{try}\:{to}\:{find}\:{it}\:{out}! \\ $$

Commented by mr W last updated on 10/Jun/25

6^2 −(R−7)^2 =R^2 −7^2 R^2 −7R−18=0 (R+2)(R−9)=0 ⇒R=9

$$\mathrm{6}^{\mathrm{2}} −\left({R}−\mathrm{7}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} \\ $$$${R}^{\mathrm{2}} −\mathrm{7}{R}−\mathrm{18}=\mathrm{0} \\ $$$$\left({R}+\mathrm{2}\right)\left({R}−\mathrm{9}\right)=\mathrm{0}\:\Rightarrow{R}=\mathrm{9} \\ $$

Commented by Tawa11 last updated on 10/Jun/25

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

Commented by Tawa11 last updated on 11/Jun/25

Sir, is there a way to know that sir is 7?

$$\mathrm{Sir},\:\mathrm{is}\:\mathrm{there}\:\mathrm{a}\:\mathrm{way}\:\mathrm{to}\:\mathrm{know}\:\mathrm{that} \\ $$$$\mathrm{sir}\:\mathrm{is}\:\:\mathrm{7}? \\ $$

Commented by fantastic last updated on 11/Jun/25

draw ⊥on 14 from the center.you will get a rectangle

$${draw}\:\bot{on}\:\mathrm{14}\:{from}\:{the}\:{center}.{you}\:{will}\:{get}\:{a}\:{rectangle} \\ $$

Commented by fantastic last updated on 11/Jun/25

$$ \\ $$

Commented by Tawa11 last updated on 11/Jun/25

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{understand}. \\ $$

Commented by mr W last updated on 11/Jun/25

$${is}\:{it}\:{not}\:{obvious}: \\ $$

Commented by mr W last updated on 11/Jun/25

Commented by Tawa11 last updated on 11/Jun/25

Yes sir, I understand it now. Thanks sir.

$$\mathrm{Yes}\:\mathrm{sir},\:\mathrm{I}\:\mathrm{understand}\:\mathrm{it}\:\mathrm{now}. \\ $$$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$

Commented by Tawa11 last updated on 11/Jun/25

Sir, please give me similar geometry. I want to practice.

$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{give}\:\mathrm{me}\:\mathrm{similar}\:\mathrm{geometry}. \\ $$$$\mathrm{I}\:\mathrm{want}\:\mathrm{to}\:\mathrm{practice}. \\ $$

Commented by mr W last updated on 11/Jun/25

$${then}\:{solve}\:{this}\:{question}\:{in}\:{a}\:{new} \\ $$$${way}!\:{i}\:{have}\:{shown}\:{three}\:{ways}. \\ $$

Commented by Tawa11 last updated on 11/Jun/25

$$\mathrm{I}\:\mathrm{will}\:\mathrm{try}\:\mathrm{sir}. \\ $$

Commented by Tawa11 last updated on 12/Jun/25

Commented by Tawa11 last updated on 12/Jun/25

$$\:\:\:\mathrm{R}^{\mathrm{2}} \:\:−\:\:\mathrm{7}^{\mathrm{2}} \:\:\:=\:\:\:\mathrm{6}^{\mathrm{2}} \:\:−\:\:\left(\mathrm{R}\:\:−\:\:\mathrm{7}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\:\mathrm{R}\:\:=\:\:\mathrm{9} \\ $$

Answered by ajfour last updated on 12/Jun/25

4r^2 −196=s^2 cos 2θ=((14)/(2r))=(7/r) sin θ=(3/r) (7/r)=1−2((3/r))^2 ⇒ r^2 −7r−18=0 r=(7/2)±(√((49+72)/4)) r= 9 units or −2 units r=9

$$\mathrm{4}{r}^{\mathrm{2}} −\mathrm{196}={s}^{\mathrm{2}} \\ $$$$\mathrm{cos}\:\mathrm{2}\theta=\frac{\mathrm{14}}{\mathrm{2}{r}}=\frac{\mathrm{7}}{{r}} \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{3}}{{r}} \\ $$$$\frac{\mathrm{7}}{{r}}=\mathrm{1}−\mathrm{2}\left(\frac{\mathrm{3}}{{r}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{r}^{\mathrm{2}} −\mathrm{7}{r}−\mathrm{18}=\mathrm{0} \\ $$$${r}=\frac{\mathrm{7}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{49}+\mathrm{72}}{\mathrm{4}}} \\ $$$${r}=\:\mathrm{9}\:{units}\:{or}\:\cancel{−\mathrm{2}\:{units}} \\ $$$$\:\:{r}=\mathrm{9} \\ $$

Commented by Tawa11 last updated on 12/Jun/25

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$

Answered by mr W last updated on 12/Jun/25

cos A=((24)/(2R))=(7/R) BD^2 =(2R)^2 −14^2 =6^2 +6^2 +2×6×6×(7/R) R^2 −((126)/R)−67=0 R^3 −67R−126=0 (R−9)(R+2)(R+7)=0 ⇒R=9

$$\mathrm{cos}\:{A}=\frac{\mathrm{24}}{\mathrm{2}{R}}=\frac{\mathrm{7}}{{R}} \\ $$$${BD}^{\mathrm{2}} =\left(\mathrm{2}{R}\right)^{\mathrm{2}} −\mathrm{14}^{\mathrm{2}} =\mathrm{6}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} +\mathrm{2}×\mathrm{6}×\mathrm{6}×\frac{\mathrm{7}}{{R}} \\ $$$${R}^{\mathrm{2}} −\frac{\mathrm{126}}{{R}}−\mathrm{67}=\mathrm{0} \\ $$$${R}^{\mathrm{3}} −\mathrm{67}{R}−\mathrm{126}=\mathrm{0} \\ $$$$\left({R}−\mathrm{9}\right)\left({R}+\mathrm{2}\right)\left({R}+\mathrm{7}\right)=\mathrm{0} \\ $$$$\Rightarrow{R}=\mathrm{9} \\ $$

Commented by Tawa11 last updated on 12/Jun/25

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

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