1. ## isosceles trapezoid help

Old guy here who's forgotten everything and needs some help.

I'm trying to make an isosceles trapezoid with a base of 26 inches and an altitude of 16 inches. The sides come in with a 6 degree pitch (84 degree angles). What I'm trying to figure out is the length of the top section that parallels the base. I don't have a clue...

Thanks!

2. Originally Posted by cluap
Old guy here who's forgotten everything and needs some help.

I'm trying to make an isosceles trapezoid with a base of 26 inches and an altitude of 16 inches. The sides come in with a 6 degree pitch (84 degree angles). What I'm trying to figure out is the length of the top section that parallels the base. I don't have a clue...

Thanks!
1. Draw a rough sketch. (see attachment)

2. $t = 26- 2x$

3. Use the tan-function in the small right triangles:

$\tan(84^\circ)=\dfrac{16}x~\implies~x=\dfrac{16}{\ tan(84^\circ)} \approx 1.681667764...$

4. Therefore $\boxed{t \approx 22.63666447...}$

3. Hello, cluap!

This requires some Trigonometry.
I hope you're prepared.

I'm trying to make an isosceles trapezoid with a base of 26 inches and an altitude of 16 inches.
The sides come in with a 6° (84° ngles).
What I'm trying to figure out is the length of the top section that parallels the base.
Code:
              B     x     C
* - - - - - *
/|           |\
/ |           | \
/6°|           |  \
/   |16         |   \
/    |           |    \
/     |           |     \
/ 84°  |           |      \
A * - - - * - - - - - * - - - * D
a   E     x     F   a
: - - - - -  26 - - - - - - :

We have isosceles trapezoid $ABCD.$
The base is: . $AD = 26.$
The height is: . $BE = 16$

Let: . $x \:=\:BC \:=\:EF$
Let: . $a \:=\:AE \:=\:FD$

Then: . $a + x + a \:=\:26 \quad\Rightarrow\quad x \:=\:26 - 2a$
. . Hence: . $BC \:=\:x \:=\:26 - 2a$ .[1]

In right triangle $BEA\!:\;\;\tan6^o \:=\:\frac{a}{16} \quad\Rightarrow\quad a \:=\:16\tan6^o \:\approx\:1.68$

Substitute into [1]: . $BC \:=\:26-2(1.68) \;=\;22.64\text{ inches}$

4. Guys - I'm sorry... I forgot to save the link, and I couldn't find this place again! Thanks for the help - you answered my questions in a way I could understand! Thanks for your help.

Take care, Paul