Re: (easy?) trapezoid question

On 9 set, 03:17, quasi <qu...@xxxxxxxx> wrote:
On Sun, 09 Sep 2007 05:00:22 -0000, "almeidabati...@xxxxxxx"

<almeidabati...@xxxxxxxxx> wrote:
Hi all! This time the situation is:

'Consider trapezioid ABCD as a right trapezoid (DCB = CBA = 90°).
Consider point M as the median point of CB (CM = DM). Consider that AM
is the bisector of angle BAD. If DC = 6 cm and AB = 10 cm, what is the
area of the trapezoid?'

This question came with a drawing, and I redid it in the computer so
that you can have a better idea of the picture:

http://img208.imageshack.us/my.php?image=trapezoidur6.jpg

Let x = the height of the trapezoid.

Let theta = angle MAD.

Then, by hypothesis, 2*theta = angle BAD.

Using triangle MAD, tan(theta) = x/20.

Using triangle BAD, tan(2*theta) = x/4.

Using the formula

tan(2*theta) = (2* tan(theta)) / (1 - tan^2(theta))

you get an equation for x.

Once you have x, you can easily compute the area.

quasi

Sorry for the late response. I understood the tan(theta) = x/20, but I
can't see where you got the tan(2*theta) from. Is BAD a right
triangle? If so, why?

PS: already solved the problem using the trapezoid median and the
Thales theorem, but this solution using tangents interests me.

.

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