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.
+1632 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
+1694 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
