Question Number 173275 by mr W last updated on 09/Jul/22
Commented by mr W last updated on 09/Jul/22
$${find}\:{the}\:{blue}\:{shaded}\:{area}=? \\ $$
Answered by mahdipoor last updated on 10/Jul/22
$${get}\:{blue}\:{is}\:{C}_{{b}} :{O}_{{b}} \left({m},{n}\right)\:,\:{r} \\ $$$${APD}\:{is}\:{C}_{\mathrm{1}} :{O}_{\mathrm{1}} \left(\mathrm{0},\mathrm{5}\right)\:,\:{r}_{\mathrm{1}} =\mathrm{5} \\ $$$${DPQC}\:{is}\:{C}_{\mathrm{2}} :{O}_{\mathrm{2}} \left(\mathrm{5},\mathrm{10}\right)\:,\:{r}_{\mathrm{2}} =\mathrm{5} \\ $$$${DQB}\:{is}\:{C}_{\mathrm{3}} ={O}_{\mathrm{3}} \left(\mathrm{0},\mathrm{0}\right)\:,\:{r}_{\mathrm{3}} =\mathrm{10} \\ $$$$\begin{cases}{\left(\mathrm{5}+{r}\right)^{\mathrm{2}} =\left({m}\right)^{\mathrm{2}} +\left(\mathrm{5}−{n}\right)^{\mathrm{2}} \:\:\:\:\:\:\left(\mathrm{1}\right)}\\{\left(\mathrm{5}−{r}\right)^{\mathrm{2}} =\left(\mathrm{5}−{m}\right)^{\mathrm{2}} +\left(\mathrm{10}−{n}\right)^{\mathrm{2}} \:\:\:\:\:\left(\mathrm{2}\right)}\\{\left(\mathrm{10}−{r}\right)^{\mathrm{2}} ={m}^{\mathrm{2}} +{n}^{\mathrm{2}} \:\:\:\:\:\:\:\left(\mathrm{3}\right)}\end{cases} \\ $$$$\Rightarrow{r}=\frac{\mathrm{40}}{\mathrm{33}}\: \\ $$
Commented by mahdipoor last updated on 10/Jul/22
$${thanks}\:{sir}\:,\:{i}\:{rechecked} \\ $$
Commented by mr W last updated on 10/Jul/22
$${thanks}\:{sir}! \\ $$
Commented by Tawa11 last updated on 11/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 09/Jul/22
Commented by mr W last updated on 10/Jul/22
![MN=5(√2) OM=5−r ON=5+r OA=10−r cos α=(((5(√2))^2 +(5+r)^2 −(5−r)^2 )/(2(5(√2))(5+r)))=((5+2r)/( (√2)(5+r))) cos β=((5^2 +(5+r)^2 −(10−r)^2 )/(2×5(5+r)))=((3r−5)/((5+r))) α+β=135° cos (α+β)=−(1/( (√2))) ((5+2r)/( (√2)(5+r)))×((3r−5)/((5+r)))−((√([2(5+r)^2 −(5+2r)^2 ]×[(5+r)^2 −(3r−5)^2 ]))/( (√2)(5+r)×(5+r)))=−(1/( (√2))) (5+2r)(3r−5)+(5+r)^2 =(√(8(25−2r^2 )(5r−r^2 ))) 15r+7r^2 =(√(8(25−2r^2 )(5r−r^2 ))) (15r+7r^2 )^2 =8(25−2r^2 )(5r−r^2 ) r(r+5)^2 (33r−40)=0 ⇒r=((40)/(33)) ✓](Q173366.png)
$${MN}=\mathrm{5}\sqrt{\mathrm{2}} \\ $$$${OM}=\mathrm{5}−{r} \\ $$$${ON}=\mathrm{5}+{r} \\ $$$${OA}=\mathrm{10}−{r} \\ $$$$\mathrm{cos}\:\alpha=\frac{\left(\mathrm{5}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left(\mathrm{5}+{r}\right)^{\mathrm{2}} −\left(\mathrm{5}−{r}\right)^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{5}\sqrt{\mathrm{2}}\right)\left(\mathrm{5}+{r}\right)}=\frac{\mathrm{5}+\mathrm{2}{r}}{\:\sqrt{\mathrm{2}}\left(\mathrm{5}+{r}\right)} \\ $$$$\mathrm{cos}\:\beta=\frac{\mathrm{5}^{\mathrm{2}} +\left(\mathrm{5}+{r}\right)^{\mathrm{2}} −\left(\mathrm{10}−{r}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{5}\left(\mathrm{5}+{r}\right)}=\frac{\mathrm{3}{r}−\mathrm{5}}{\left(\mathrm{5}+{r}\right)} \\ $$$$\alpha+\beta=\mathrm{135}° \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\frac{\mathrm{5}+\mathrm{2}{r}}{\:\sqrt{\mathrm{2}}\left(\mathrm{5}+{r}\right)}×\frac{\mathrm{3}{r}−\mathrm{5}}{\left(\mathrm{5}+{r}\right)}−\frac{\sqrt{\left[\mathrm{2}\left(\mathrm{5}+{r}\right)^{\mathrm{2}} −\left(\mathrm{5}+\mathrm{2}{r}\right)^{\mathrm{2}} \right]×\left[\left(\mathrm{5}+{r}\right)^{\mathrm{2}} −\left(\mathrm{3}{r}−\mathrm{5}\right)^{\mathrm{2}} \right]}}{\:\sqrt{\mathrm{2}}\left(\mathrm{5}+{r}\right)×\left(\mathrm{5}+{r}\right)}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\left(\mathrm{5}+\mathrm{2}{r}\right)\left(\mathrm{3}{r}−\mathrm{5}\right)+\left(\mathrm{5}+{r}\right)^{\mathrm{2}} =\sqrt{\mathrm{8}\left(\mathrm{25}−\mathrm{2}{r}^{\mathrm{2}} \right)\left(\mathrm{5}{r}−{r}^{\mathrm{2}} \right)} \\ $$$$\mathrm{15}{r}+\mathrm{7}{r}^{\mathrm{2}} =\sqrt{\mathrm{8}\left(\mathrm{25}−\mathrm{2}{r}^{\mathrm{2}} \right)\left(\mathrm{5}{r}−{r}^{\mathrm{2}} \right)} \\ $$$$\left(\mathrm{15}{r}+\mathrm{7}{r}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{8}\left(\mathrm{25}−\mathrm{2}{r}^{\mathrm{2}} \right)\left(\mathrm{5}{r}−{r}^{\mathrm{2}} \right) \\ $$$${r}\left({r}+\mathrm{5}\right)^{\mathrm{2}} \left(\mathrm{33}{r}−\mathrm{40}\right)=\mathrm{0} \\ $$$$\Rightarrow{r}=\frac{\mathrm{40}}{\mathrm{33}}\:\checkmark \\ $$
Commented by Tawa11 last updated on 11/Jul/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$