1. ## Triangle. Find QT

Okay so Ive figured out half of this question so far, and I need help finding out the rest.

PQRS is an isosceles trapezoid with PS=QR=17 and PQ = 16 and SR = 32. Point R is on SV with RV = 16 and point Q is on the hypotenue ST of right triangle STV. The length of QT is:
a) 26 b) 3 x root of 89 c) 4 x root of 57 d) 3 x root of 93 e) 39

So Ive figured out that the height of the trapezoid is 15
and that QS = root of 801
can someone help me find QT with this information given and explain why ?

2. Hello, foreverbrokenpromises!

I can't get your diagram to show up
. . but I think I've figured it out.

$\displaystyle PQRS$ is an isosceles trapezoid with: .$\displaystyle PS=QR=17,\;PQ = 16,\;SR = 32$

$\displaystyle SR$ is extended to $\displaystyle V$ so that $\displaystyle RV = 16.$

Point $\displaystyle Q$ is on the hypotenue $\displaystyle ST$ of right triangle $\displaystyle STV.$

The length of $\displaystyle QT$ is:

. . $\displaystyle (a)\;26\qquad (b)\;3\sqrt{89} \qquad (c)\;4\sqrt{57} \qquad (d)\;3\sqrt{93}\qquad (e)\;39$

So I've figured out that the height of the trapezoid is 15 . Right!
. . and that $\displaystyle QS \:=\: \sqrt{801} \:=\:3\sqrt{89}$ . Yes!

Can someone help me find $\displaystyle QT$ with this information given and explain why?
Code:
                                          o T
*  |
*     |
*        |
*           |
P    16     Q  *              |
o-----------o                 |
/:        *  :\                |
/ :     *     : \               |
/  :  *        :  \              |
/   *         15:   \             |
/ *  :           :    \            |
o-----+-----------+-----o - - - - - o
S  8  A    16     B  8  R    16     V

Draw $\displaystyle PA \perp SR$
Draw $\displaystyle QB \perp SR$
Then: .$\displaystyle SA = 8,\;AB = 16,\;BR = 8$

We see that: .$\displaystyle \Delta QBS \sim \Delta TVS$

Hence: .$\displaystyle \frac{ST}{SQ} \:=\:\frac{SV}{SB} \quad\Rightarrow\quad \frac{ST}{SQ} \:=\:\frac{48}{24} \quad\Rightarrow\quad ST \:=\:2\!\cdot\!QS$

Divide by 2: . $\displaystyle \tfrac{1}{2}ST \:=\:QS$

Since $\displaystyle QT = \tfrac{1}{2}ST$, we have: .$\displaystyle QT \:=\:QS \:=\:3\sqrt{89}$ . . . answer (b)

3. ## Here is my answer.

Some how my answer is not one of the choice. I am pretty sure of my answer, but I could be wrong. Please let me know.

QT=34

I know what I did wrong! Sorry!