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In triangle ABC, AB = 17, AC = 8, and BC = 20. Let D be the foot of the altitude from C to AB. Find the area of triangle ACD.

 
 Aug 5, 2022

Best Answer 

 #1
avatar+13793 
+2

 Find the area of triangle ACD.

 

Hello Guest!

 

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

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

laugh  !

 
 Aug 5, 2022
edited by asinus  Aug 5, 2022
edited by asinus  Aug 5, 2022
 #1
avatar+13793 
+2
Best Answer

 Find the area of triangle ACD.

 

Hello Guest!

 

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

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

laugh  !

 
asinus Aug 5, 2022
edited by asinus  Aug 5, 2022
edited by asinus  Aug 5, 2022
 #2
avatar+124525 
+1

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 

 

 

cool cool cool 

 
 Aug 6, 2022
edited by CPhill  Aug 6, 2022
 #3
avatar+113 
+1
 Aug 6, 2022
edited by tuffla2022  Aug 6, 2022
edited by tuffla2022  Aug 6, 2022

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