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.

Guest Feb 9, 2022

#1**-1 **

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**

proyaop Feb 10, 2022

#2**+1 **

AD => **x** BD => **17 - x**

8^{2} - x^{2} = 16^{2} - (17 - x)^{2}

**x = 97/34 AD = 97/34**

CD = sqrt(8^{2} - x^{2}) **CD = (15√287) / 34**

**[ACD] = 1/2(AD * CD) **

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

civonamzuk Feb 10, 2022

edited by
Guest
Feb 10, 2022