Posted by Denis Borris on October 27, 2002 at 20:06:00:
In Reply to: Re: right triangles posted by Joel on October 27, 2002 at 18:31:33:
....if you don't like using "slopes" (like me!).
Move the points 1 "up": (3,6), (5,0) and (k,9);
makes for easier and clearer graphing.
Label (3,6) as A, (5,0) as B
Using A,B form right triangle ABC (angle C = 90 degrees)
So AC = 6 and BC = 2.
From A, draw line AD perpendicular to AB, D on line y=9;
using A,D form right triangle ADE (angle E = 90 degrees)
Now you have triangle ADE similar to triangle ABC; ED
easily calculated to equal 9, hence "k" = 12.
And right triangle ADB has been formed.
Repeat above with a perpendicular from B; again, you'll
get similar triangles, with "k" ending up at 30.
These will be the only 2 values of "k".
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