Question Number 181673 by cherokeesay last updated on 28/Nov/22
![](https://www.tinkutara.com/question/27981.png)
Answered by HeferH last updated on 28/Nov/22
![A = (((r_1 + r_2 )BD)/2) BD = 5 − (((16)/5) + 1) = ((25 − 21)/5) = (4/5) A = (1/2)(2 + ((12)/5))(4/5) = ((44)/(25)) u^2](https://www.tinkutara.com/question/Q181708.png)
Commented by cherokeesay last updated on 29/Nov/22
![Nice solution, thank you sir.](https://www.tinkutara.com/question/Q181726.png)
Answered by a.lgnaoui last updated on 29/Nov/22
![Area=1,76](https://www.tinkutara.com/question/Q181713.png)
Answered by a.lgnaoui last updated on 29/Nov/22
![](https://www.tinkutara.com/question/27995.png)
Answered by a.lgnaoui last updated on 29/Nov/22
![tan α=(3/4) α=36,87 AE^2 =2^2 +4^2 AE=2(√5) ⇒DE=4 EF=(√(CE^2 +CF^2 )) =(√(16+9)) =5 sin α=((CF)/(EF))=(3/5) =((BC)/(CE)) BC=((12)/5)=2,4 BE^2 =CE^2 −BC^2 ⇒ BE=3,2 alors BD=DE−BE=0,8 par consequant: Aire=A_1 +A_2 A_1 =(1/2)(AD×BD)=((2×0,8)/2)=0,8 A_2 =(1/2)(BD×BC)=((2,4×0,8)/2)=0,96 A=1,76](https://www.tinkutara.com/question/Q181715.png)
Commented by cherokeesay last updated on 29/Nov/22
![Nice ! thank you so much.](https://www.tinkutara.com/question/Q181727.png)