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Question-173275




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=?
$${find}\:{the}\:{blue}\:{shaded}\:{area}=? \\ $$
Answered by mahdipoor last updated on 10/Jul/22
get blue is C_b :O_b (m,n) , r  APD is C_1 :O_1 (0,5) , r_1 =5  DPQC is C_2 :O_2 (5,10) , r_2 =5  DQB is C_3 =O_3 (0,0) , r_3 =10   { (((5+r)^2 =(m)^2 +(5−n)^2       (1))),(((5−r)^2 =(5−m)^2 +(10−n)^2      (2))),(((10−r)^2 =m^2 +n^2        (3))) :}  ⇒r=((40)/(33))
$${get}\:{blue}\:{is}\:{C}_{{b}} :{O}_{{b}} \left({m},{n}\right)\:,\:{r} \\ $$$${APD}\:{is}\:{C}_{\mathrm{1}} :{O}_{\mathrm{1}} \left(\mathrm{0},\mathrm{5}\right)\:,\:{r}_{\mathrm{1}} =\mathrm{5} \\ $$$${DPQC}\:{is}\:{C}_{\mathrm{2}} :{O}_{\mathrm{2}} \left(\mathrm{5},\mathrm{10}\right)\:,\:{r}_{\mathrm{2}} =\mathrm{5} \\ $$$${DQB}\:{is}\:{C}_{\mathrm{3}} ={O}_{\mathrm{3}} \left(\mathrm{0},\mathrm{0}\right)\:,\:{r}_{\mathrm{3}} =\mathrm{10} \\ $$$$\begin{cases}{\left(\mathrm{5}+{r}\right)^{\mathrm{2}} =\left({m}\right)^{\mathrm{2}} +\left(\mathrm{5}−{n}\right)^{\mathrm{2}} \:\:\:\:\:\:\left(\mathrm{1}\right)}\\{\left(\mathrm{5}−{r}\right)^{\mathrm{2}} =\left(\mathrm{5}−{m}\right)^{\mathrm{2}} +\left(\mathrm{10}−{n}\right)^{\mathrm{2}} \:\:\:\:\:\left(\mathrm{2}\right)}\\{\left(\mathrm{10}−{r}\right)^{\mathrm{2}} ={m}^{\mathrm{2}} +{n}^{\mathrm{2}} \:\:\:\:\:\:\:\left(\mathrm{3}\right)}\end{cases} \\ $$$$\Rightarrow{r}=\frac{\mathrm{40}}{\mathrm{33}}\: \\ $$
Commented by mahdipoor last updated on 10/Jul/22
thanks sir , i rechecked
$${thanks}\:{sir}\:,\:{i}\:{rechecked} \\ $$
Commented by mr W last updated on 10/Jul/22
thanks sir!
$${thanks}\:{sir}! \\ $$
Commented by Tawa11 last updated on 11/Jul/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$
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)) ✓
$${MN}=\mathrm{5}\sqrt{\mathrm{2}} \\ $$$${OM}=\mathrm{5}−{r} \\ $$$${ON}=\mathrm{5}+{r} \\ $$$${OA}=\mathrm{10}−{r} \\ $$$$\mathrm{cos}\:\alpha=\frac{\left(\mathrm{5}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left(\mathrm{5}+{r}\right)^{\mathrm{2}} −\left(\mathrm{5}−{r}\right)^{\mathrm{2}} }{\mathrm{2}\left(\mathrm{5}\sqrt{\mathrm{2}}\right)\left(\mathrm{5}+{r}\right)}=\frac{\mathrm{5}+\mathrm{2}{r}}{\:\sqrt{\mathrm{2}}\left(\mathrm{5}+{r}\right)} \\ $$$$\mathrm{cos}\:\beta=\frac{\mathrm{5}^{\mathrm{2}} +\left(\mathrm{5}+{r}\right)^{\mathrm{2}} −\left(\mathrm{10}−{r}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{5}\left(\mathrm{5}+{r}\right)}=\frac{\mathrm{3}{r}−\mathrm{5}}{\left(\mathrm{5}+{r}\right)} \\ $$$$\alpha+\beta=\mathrm{135}° \\ $$$$\mathrm{cos}\:\left(\alpha+\beta\right)=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\frac{\mathrm{5}+\mathrm{2}{r}}{\:\sqrt{\mathrm{2}}\left(\mathrm{5}+{r}\right)}×\frac{\mathrm{3}{r}−\mathrm{5}}{\left(\mathrm{5}+{r}\right)}−\frac{\sqrt{\left[\mathrm{2}\left(\mathrm{5}+{r}\right)^{\mathrm{2}} −\left(\mathrm{5}+\mathrm{2}{r}\right)^{\mathrm{2}} \right]×\left[\left(\mathrm{5}+{r}\right)^{\mathrm{2}} −\left(\mathrm{3}{r}−\mathrm{5}\right)^{\mathrm{2}} \right]}}{\:\sqrt{\mathrm{2}}\left(\mathrm{5}+{r}\right)×\left(\mathrm{5}+{r}\right)}=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$$\left(\mathrm{5}+\mathrm{2}{r}\right)\left(\mathrm{3}{r}−\mathrm{5}\right)+\left(\mathrm{5}+{r}\right)^{\mathrm{2}} =\sqrt{\mathrm{8}\left(\mathrm{25}−\mathrm{2}{r}^{\mathrm{2}} \right)\left(\mathrm{5}{r}−{r}^{\mathrm{2}} \right)} \\ $$$$\mathrm{15}{r}+\mathrm{7}{r}^{\mathrm{2}} =\sqrt{\mathrm{8}\left(\mathrm{25}−\mathrm{2}{r}^{\mathrm{2}} \right)\left(\mathrm{5}{r}−{r}^{\mathrm{2}} \right)} \\ $$$$\left(\mathrm{15}{r}+\mathrm{7}{r}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{8}\left(\mathrm{25}−\mathrm{2}{r}^{\mathrm{2}} \right)\left(\mathrm{5}{r}−{r}^{\mathrm{2}} \right) \\ $$$${r}\left({r}+\mathrm{5}\right)^{\mathrm{2}} \left(\mathrm{33}{r}−\mathrm{40}\right)=\mathrm{0} \\ $$$$\Rightarrow{r}=\frac{\mathrm{40}}{\mathrm{33}}\:\checkmark \\ $$
Commented by Tawa11 last updated on 11/Jul/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

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