1. ## Right Triangles

Find the value of x and y:

2. Originally Posted by icebreaker09
Find the value of x and y:

I'm sorry, but I find this to be a bit ambiguous. Which segments are 5, 12, and 13? (I've got what I think is a good guess, but I need to be sure.)

-Dan

3. Sorry, but that's all it said: I can post the first part though (already solved it): A frame in the shape of the simple scissors truss shown a the right below can be used to support a peaked roof. The weight of the roof compresses some parts of the frame, while other parts are in tension. A frame made with s segments joined at j points is stable is s is greater to or less than 2j-3. In this truss, 9 segments connect 6 points. Verify that the truss is stable. Then find the values of x and y.

4. For convenience I'm going to label the points top down and left to right with A, B, C, etc.

There are two right triangles: triangle BDE and CDG. The right angles are angles DBE and DCG, though the diagram doesn't make it look like that.

The problem I am having here is that the apex angle EAG is not specified. We may have a number of possible apex angles for this construction, all of which give different x and y values.

-Dan

5. Hello, icebreaker!

Find the value of $x$ and $y$.
Code:
                A
*
/*\
/ * \
y /  *  \ y
/   *   \
/    *    \
/     *     \
F *  5   *   5  * E
/   *   *   *   \
12 /        O        \ 12
/  13 *   *   * 13  \
/  *       *       *  \
B * - - - - - * - - - - - * C
x     D     x

Triangles OFB and OEC have sides: 5, 12, 13.
. . Since $5^2 + 12^2 \,=\,13^2$, they are right triangles.
Then: $\angle OFB = \angle OEC = 90^o$

In right triangle $BEC\!:\;\;BC^2 \:=\:BE^2 + EC^2$
. . That is: . $(2x)^2\:=\:18^2+12^2\quad\Rightarrow\quad4x^2\,=\, 468\quad\Rightarrow\quad x^2\,=\,117$
Therefore: . $x \:=\:\sqrt{117}\quad\Rightarrow\quad\boxed{x \,=\,3\sqrt{13}}$

In right triangle $BEA\!:\;\;FA^2 \:=\:BE^2 + AE^2$
. . That is: . $(y+12)^2\:=\:18^2 + y^2\quad\Rightarrow\quad y^2 + 24y + 144\:=\:324 + y^2$
Therefore: . $24y \,=\,180\quad\Rightarrow\quad\boxed{y \,=\,\frac{15}{2}}$

6. Originally Posted by Soroban
Hello, icebreaker!

Triangles OFB and OEC have sides: 5, 12, 13.
. . Since $5^2 + 12^2 \,=\,13^2$, they are right triangles.
Then: $\angle OFB = \angle OEC = 90^o$

In right triangle $BEC\!:\;\;BC^2 \:=\:BE^2 + EC^2$
. . That is: . $(2x)^2\:=\:18^2+12^2\quad\Rightarrow\quad4x^2\,=\, 468\quad\Rightarrow\quad x^2\,=\,117$
Therefore: . $x \:=\:\sqrt{117}\quad\Rightarrow\quad\boxed{x \,=\,3\sqrt{13}}$

In right triangle $BEA\!:\;\;FA^2 \:=\:BE^2 + AE^2$
. . That is: . $(y+12)^2\:=\:18^2 + y^2\quad\Rightarrow\quad y^2 + 24y + 144\:=\:324 + y^2$
Therefore: . $24y \,=\,180\quad\Rightarrow\quad\boxed{y \,=\,\frac{15}{2}}$

Am I being too suspiscious here? How do we know that BO and OE form a straight line when connected?

-Dan