# Geometry help

• Mar 8th 2009, 04:36 PM
Itachi888Uchiha
Geometry help
Find the altitude (length of a segment perpendicular to both bases) of an isosceles trapezoid with sides 9,12,9,18.

How do I solve?
• Mar 8th 2009, 05:02 PM
skeeter
Quote:

Originally Posted by Itachi888Uchiha
Find the altitude (length of a segment perpendicular to both bases) of an isosceles trapezoid with sides 9,12,9,18.

How do I solve?

did you make a sketch? draw the height from a corner where the 9 and 12 sides meet down to the longer base (18). notice the right triangle?

use Pythagoras ...

$\displaystyle h = \sqrt{9^2 - 3^2}$
• Mar 8th 2009, 05:08 PM
Mentia
One way to do it is to make a right triangle along one of the sides:

An isosceles trapezoid looks like this (by mathwords.com):

http://www.mathwords.com/i/i_assets/i121.gif

So in this case our long base is 18, short base is 12. If we draw a straight line down from the corner of the small base and one of the legs, we can get the height of the resulting triangle which will be the same as the height of the trapezoid.

We know that the long base is 6 longer than the short base, so the base of our triangle must be half of that, or 3. Thus we have a right triangle with base 3 and hypotenuse 9. We can use the pythagorean theorem to get the height.

9^2 = 3^2 - (height)^2

height = sqrt(81 - 9) = about 8.5 = altitude
• Mar 8th 2009, 05:10 PM
Itachi888Uchiha
Thanks. I thought I had to draw an altitude at the middle of the trapezoid, and I was getting confused. I got 6 root 2 by the way....
EDIT: I can leave my answers in the form of a radical if I want......