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# triangles

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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

#1
+2

Find the area of triangle ACD.

Hello Guest!

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

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

Aug 5, 2022
edited by asinus  Aug 5, 2022
edited by asinus  Aug 5, 2022

#1
+2

Find the area of triangle ACD.

Hello Guest!

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

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

asinus Aug 5, 2022
edited by asinus  Aug 5, 2022
edited by asinus  Aug 5, 2022
#2
+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   Aug 6, 2022
edited by CPhill  Aug 6, 2022
#3
+1

My go at this one: https://www.geogebra.org/classic/zbqaxepk

Aug 6, 2022
edited by tuffla2022  Aug 6, 2022
edited by tuffla2022  Aug 6, 2022