# Find the possible values of X using "Similar right triangles"?

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• Feb 26th 2011, 05:44 AM
Lizarrdo
Find the possible values of X using "Similar right triangles"?
Find the possible values of x.

Attachment 20964
I got 6, but the answer is 12. Why is that? My teachers said there was an easy way to do it without quadratics but I don't remember.
• Feb 26th 2011, 06:20 AM
BAdhi
there are three right angled triangles that you can use to apply phythagoras theorem. apply that theorem to two of them find the lenght of unknown sides with the variable x.

then apply the pythagoras theorem to the other triangle to find x
• Feb 26th 2011, 06:24 AM
Sambit
If you want to avoid quadratics, try to use the fact that the sides of the opposite of the same angles of two similar triangles are proportional in length.
• Feb 26th 2011, 06:30 AM
Lizarrdo
Quote:

Originally Posted by BAdhi
there are three right angled triangles that you can use to apply phythagoras theorem. apply that theorem to two of them find the lenght of unknown sides with the variable x.

then apply the pythagoras theorem to the other triangle to find x

So

(x-3)^2 + x^2 = x^2

x^2 - 9 + x^2 = x^2

2x^2 - 9 = x^2

-9 = -x^2

x=3?

I did something wrong for sure. How would I do teh square root of (x-3)?
• Feb 26th 2011, 06:31 AM
Soroban
Hello, Lizarrdo!

Quote:

$\text}Find the possible values of }x.$

Code:

            C             *           *| *           * |  *  x         *  |h    *         *  |      *       *    |        *     A *  *  *  *  *  *  * B       x-5  D    x-3

Note that $x > 5.$

We have right triangle $ABC,\:\angle C = 90^o.$

The altitude to the hypotenuse is: . $h = CD.$
It divides the hypotenuse into two segments: . $AD = x-5,\;DB = x-3.$

Fact: The altitude to the hypotenuse is the geometric mean
. . . . . . . . . of the segments of the hypotenuse.

Hence, we have: . $h^2 \:=\:(x-3)(x-5)$ .[1]

In right triangle $CDB\!:\;h^2 + (x-3)^2 \:=\:x^2 \quad\Rightarrow\quad h^2 \:=\:x^2-(x-3)^2$ .[2]

Equate [1] and [2]: . $(x-5)(x-3) \:=\:x^2 - (x-3)^2$

We have: . $x^2 -14x + 24 \:=\:0 \quad\Rightarrow\quad (x-2)(x-12) \:=\:0$

. . Therefore: . $x \,=\,12$