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In triangle ABC, AB = 17, AC = 8, and BC =  \(16\). Let D be the foot of the altitude from C to AB. Find the area of triangle ACD.

 Feb 9, 2022
 #1
avatar+514 
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First I gave CD a variable, h. AD as x, and DB as y.

Here our some equations:

\(x^2 + h^2 = 64\)

\(y^2+h^2 = 256\)

\(x+y=17\)

 

Now we can solve for x. 

h^2 = -(x^2) + 64

Plugging that into the second equation we have:

y^2 - x^2 + 64 = 256

y^2 - x^2 = 192.

 

Using difference of squares, we have:

(x + y)(y - x) = 192

 

Thus 17y - 17x = 192.

We also can multiply equation 3 by 17, so we have 17y + 17x = 289.

 

Then adding the subtracting the two equations from each other we have -34x = -97.

x = 97/34.

 

Using the pythagorean theorem, we get CD as 5sqrt(2579), so the area of the triangle is 485sqrt(2579)/34

 

indecision

 Feb 10, 2022
 #2
avatar+1696 
+1

AD => x          BD => 17 - x

 

82 - x2 = 162 - (17 - x)2

 

x = 97/34                 AD = 97/34

 

CD = sqrt(82 - x2)       CD = (15√287) / 34

 

[ACD] = 1/2(AD * CD) 

 

[ACD] = (1455√287) / 2312 = 10.66144601

 Feb 10, 2022
edited by Guest  Feb 10, 2022

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