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Question-223851

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Question Number 223851 by Tawa11 last updated on 06/Aug/25
Answered by mr W last updated on 07/Aug/25
Commented by mr W last updated on 08/Aug/25
geometry expressions
$${geometry}\:{expressions} \\ $$
Commented by mr W last updated on 07/Aug/25
(6+r)^2 =10^2 +r^2 ⇒r=((16)/3) cos α=((10)/(6+((16)/3)))=((15)/(17)), sin α=(((16)/3)/(6+((16)/3)))=(8/(17)) ((DB)/(10))=((r+12)/(6+r)) ⇒DB=(1+(6/(6+((16)/3))))×10=((260)/(17)) EB^2 =(10+((260)/(17)))^2 +(6+2×((16)/3))^2 −2(10+((260)/(17)))(6+2×((16)/3))×((15)/(17)) ⇒EB=((20(√(1129)))/(51)) sin β=(((6+3×((16)/3))×(8/(17)))/((20(√(1129)))/(51)))=((20)/( (√(1129)))) γ=(π/2)+α=(π/2)+sin^(−1) (8/(17)) shaded area =(1/2)×((260)/(17))×((20(√(1129)))/(51))×((20)/( (√(1129))))+(1/2)×(((16)/3))^2 (((15)/(17))−(π/2)−sin^(−1) (8/(17))) =((20960)/( 289))−((128)/9)((π/2)+sin^(−1) (8/(17))) ≈43.217455
$$\left(\mathrm{6}+{r}\right)^{\mathrm{2}} =\mathrm{10}^{\mathrm{2}} +{r}^{\mathrm{2}} \\ $$$$\Rightarrow{r}=\frac{\mathrm{16}}{\mathrm{3}} \\ $$$$\mathrm{cos}\:\alpha=\frac{\mathrm{10}}{\mathrm{6}+\frac{\mathrm{16}}{\mathrm{3}}}=\frac{\mathrm{15}}{\mathrm{17}},\:\mathrm{sin}\:\alpha=\frac{\frac{\mathrm{16}}{\mathrm{3}}}{\mathrm{6}+\frac{\mathrm{16}}{\mathrm{3}}}=\frac{\mathrm{8}}{\mathrm{17}} \\ $$$$\frac{{DB}}{\mathrm{10}}=\frac{{r}+\mathrm{12}}{\mathrm{6}+{r}} \\ $$$$\Rightarrow{DB}=\left(\mathrm{1}+\frac{\mathrm{6}}{\mathrm{6}+\frac{\mathrm{16}}{\mathrm{3}}}\right)×\mathrm{10}=\frac{\mathrm{260}}{\mathrm{17}} \\ $$$${EB}^{\mathrm{2}} =\left(\mathrm{10}+\frac{\mathrm{260}}{\mathrm{17}}\right)^{\mathrm{2}} +\left(\mathrm{6}+\mathrm{2}×\frac{\mathrm{16}}{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{10}+\frac{\mathrm{260}}{\mathrm{17}}\right)\left(\mathrm{6}+\mathrm{2}×\frac{\mathrm{16}}{\mathrm{3}}\right)×\frac{\mathrm{15}}{\mathrm{17}} \\ $$$$\Rightarrow{EB}=\frac{\mathrm{20}\sqrt{\mathrm{1129}}}{\mathrm{51}} \\ $$$$\mathrm{sin}\:\beta=\frac{\left(\mathrm{6}+\mathrm{3}×\frac{\mathrm{16}}{\mathrm{3}}\right)×\frac{\mathrm{8}}{\mathrm{17}}}{\frac{\mathrm{20}\sqrt{\mathrm{1129}}}{\mathrm{51}}}=\frac{\mathrm{20}}{\:\sqrt{\mathrm{1129}}} \\ $$$$\gamma=\frac{\pi}{\mathrm{2}}+\alpha=\frac{\pi}{\mathrm{2}}+\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{8}}{\mathrm{17}} \\ $$$${shaded}\:{area}\:=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{260}}{\mathrm{17}}×\frac{\mathrm{20}\sqrt{\mathrm{1129}}}{\mathrm{51}}×\frac{\mathrm{20}}{\:\sqrt{\mathrm{1129}}}+\frac{\mathrm{1}}{\mathrm{2}}×\left(\frac{\mathrm{16}}{\mathrm{3}}\right)^{\mathrm{2}} \left(\frac{\mathrm{15}}{\mathrm{17}}−\frac{\pi}{\mathrm{2}}−\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{8}}{\mathrm{17}}\right) \\ $$$$\:\:\:\:\:=\frac{\mathrm{20960}}{\:\mathrm{289}}−\frac{\mathrm{128}}{\mathrm{9}}\left(\frac{\pi}{\mathrm{2}}+\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{8}}{\mathrm{17}}\right) \\ $$$$\:\:\:\:\:\approx\mathrm{43}.\mathrm{217455} \\ $$
Commented by Tawa11 last updated on 07/Aug/25
Thanks sir. I really appreciate.
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$
Commented by mr W last updated on 07/Aug/25
Commented by Tawa11 last updated on 08/Aug/25
Wow, great sir.
$$\mathrm{Wow},\:\mathrm{great}\:\mathrm{sir}. \\ $$
Commented by Tawa11 last updated on 08/Aug/25
What is the name of this app sir.
$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{name}\:\mathrm{of}\:\mathrm{this}\:\mathrm{app}\:\mathrm{sir}. \\ $$
Commented by fantastic last updated on 08/Aug/25
sir can you please provide the link?
$${sir}\:{can}\:{you}\:{please}\:{provide}\:{the}\:{link}? \\ $$
Commented by fantastic last updated on 08/Aug/25
I think it is premium
$${I}\:{think}\:{it}\:{is}\:{premium} \\ $$
Commented by MathematicalUser2357 last updated on 11/Aug/25
You can buy a "premium" app after installing the app. The "premium" app is not instantly installed. And you can buy an "expensive" app instantly. You can say: I think it is expensive

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