isosceles trapezoid help

cluap · #1 $Old$ October 19th, 2009, 10:56 PM

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!

earboth · #2 $Old$ October 20th, 2009, 02:00 AM

Quote:

Originally Posted by cluap $View Post$

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

Soroban · #3 $Old$ October 20th, 2009, 07:49 AM

Hello, cluap!

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

Quote:

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

cluap · #4 $Old$ November 2nd, 2009, 07:10 PM

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