I am trying to teach myself trig, but this problem has me stumped:
One side of a triangle is 18 cm.
Inside the triangle, a segment
is drawn parallel to the 18-cm.
side, forming a trapezoid whose
area is 1/3 that of the original
triangle. Find the length of the
segment.
Any answers or at least clues?
The triangle outside of the trapezoid is similar to the original triangle. Between that and the standard formulae for areas of triangles and trapezoids, you should be good to go.
Here is a trig site that I find pretty cool.
I guess it’s java based so you can grab a point (such as B) and move it around and see what moves what where.
I picked the sine page just for convenience sake, it’s a whole series covering many aspects with the java thing on each.
I wish we had that kind of interactive thing back in the day instead of listening to Dr. Turtlemann drone on and on.
First thing you should do is draw a picture.
Your picture should look like a big triangle with a line segment drawn inside it. The line spilts the big triangle into a little triangle and a trapezoid.
The little triangle has 2/3 the area of the big triangle (since you’re told the trapezoid takes up the other 1/3 the area). It’s also similar to the big triangle, meaning it’s the same triangle except its dimensions are scaled down.
The area of a triangle is A = 1/2 * base * height.
In order to shrink the triangle while keeping it proportionate to the original, you shrink both the base and height by the same factor. If we want to reduce the total area to 2/3 the original, we multiply both the base and the height by the square root of 2/3.
Thus, the base of the small triangle is 18 cm * sqrt(2/3) = 14.6969
Does this solution make sense to you? If part is unclear I can explain it more thoroughly.
I do not understand the Square root part. If I do not use the square root I get a nice even number, 12, and if I do it is some weird decimal number. It seems like the answer is most likely a simple number.
The area of a triangle is 1/2 x base x height.
If you multiply both the base and height by k to produce a similarly shaped triangle, the new area is 1/2 x (k x base) x (k x height). Put it another way, 1/2 x base x height x k[sup]2[/sup]. That is, the new area is the original area multiplied by k[sup]2[/sup].
So if you want the new area to be 2/3 of the original area (i.e. you want k[sup]2[/sup] to be 2/3), k must be the square root of 2/3.
That’s the Math Student’s Fallacy 
Suppose the answer is 12. The area of the original triangle is
18h/2
where h is the height (we don’t know what it is). So the area of the whole triangle is 9h.
If we draw a line parallel to the base of length 12, that is 2/3 of 18. Therefore the height of the triangle that has base 12 is (2/3)h (we know this because they are similar triangles, mentioned above). So the area of the smaller triangle is
12(2/3)h/2
or 4h. Now, we know that, by the problem statement, the trapezoid is 1/3 of the large triangle, therefore the small triangle is 2/3 the area of the larger one. But 4h is not 2/3 of 9h. So the answer can’t be 12.
(By the way, this is plane geometry, not trig.)
You want the area of the small triangle to be 2/3 the area of the large triangle.
You also know they have to be similar. (You can see this by using the various theorems of geometry to recognize that the angles of the small triangle are the same as the angles of the large triangle.) Basically, similar means proportionate. If we’re shrinking down the large triangle to get the small triangle, we have to shrink all of its dimensions by the same factor.
If we multiply the base of the large triangle by X, and we multiply the height by X, then we’ve multiplied the area by X[sup]2[/sup]. (This follows from Area = 1/2 * base*height)
So, to get the right change in area from the large triangle to the small one, we need X[sup]2[/sup] = 2/3
Unfortunately in math not everything works out to nice round numbers. If I told you the solution to X[sup]2[/sup] = 2 was X = 1.41421356 would you say “No way, that’s some weird decimal instead of a nice round number!”? That’s just the way it is.
As others have said, you actually don’t need any trig for this problem, just ordinary high school geometry.