triangles

 Find the area of triangle ACD.

Hello Guest!

$f(x)=\sqrt{8^2-x^2}\\ g(x)=\sqrt{20^2-(x-17)^2}\\ \sqrt{8^2-x^2}=\sqrt{20^2-(x-17)^2}\\64-x^2=400-x^2+34x-289\\ 34x=-47\\ x=-\dfrac{47}{34}\\ y=\sqrt{8^2-(-\frac{47}{34})^2}$

$A_{ACD}=\dfrac{xy}{2}\\ A_{ACD}=\dfrac{1.382\cdot 7.8985}{2}\\ \color{blue}A_{ACD}=5.469$

  !

This triangle is obtuse.. altitude CD falls outside triangle ABC  as shown

Let x be the distance from A to D

We have that

DC^2 = BC^2 +  (AB + x)^2 =   20^2  - (17+x)^2

DC^2 = (AC)^2  - x^2  =    8^2  - x^2

So

20^2 - (17+x)^2 = 8^2 - x^2

400 - x^2 - 34x -289  = 64 - x^2

-34x  =  289 + 64 - 400

-34x = -47

x = 47/34

DC =  sqrt [8^2 - (47/34)^2 ]

Area of ACD =   (1/2) (AD) (CD)  =      (1/2)(47/34)(sqrt [ 8^2 - (47/34)^2 ]   =   5.446 

My go at this one:

https://www.geogebra.org/classic/zbqaxepk