Hello, dearlybelovedx3!
In right triangle ABC, BD is the altitude to hypotenuse AC.
$\displaystyle AC = 20,\;BD = 6,\;AD < DC$
(a) Find $\displaystyle AD$ and $\displaystyle DC.$
(b) To the nearest tenth, find $\displaystyle AB.$ Code:
B
*
*| *
* | *
* |6 *
* | *
* | *
A * * * + * * * * * * * * * C
x D 20-x
: - - - - 20 - - - - - :
We have three similar right triangles: .$\displaystyle \Delta ABC \sim \Delta BDC \sim \Delta ADB$
Let $\displaystyle AD\,=\,x$, then $\displaystyle DC \,=\,20-x$
Then: .$\displaystyle \frac{x}{6} \:=\:\frac{6}{20-x}\quad\Rightarrow\quad x^2-20x+36\:=\:0$
. . and we have: .$\displaystyle (x-2)(x-18)\:=\:\quad\Rightarrow\quad x \:=\:2,\:18$
(a) Therefore: .$\displaystyle AD \,=\,2,\;DC\,=\,18$
(b) In right triangle ADB, we have: .$\displaystyle AB^2 \:=\:2^2 + 6^2 \quad\Rightarrow\quad AB \:=\:\sqrt{40}\:\approx\:6.3$