# [SOLVED] what is the answer to this problem

• Nov 25th 2007, 06:43 PM
dearlybelovedx3
[SOLVED] what is the answer to this problem
• Nov 25th 2007, 07:45 PM
Soroban
Hello, dearlybelovedx3!

Quote:

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: .$\displaystyle AB^2 \:=\:2^2 + 6^2 \quad\Rightarrow\quad AB \:=\:\sqrt{40}\:\approx\:6.3$