Inscribed triangle

Hey SzDave,

Can you show me your method? That should have gotten you the right answer. This has to be a right triangle, because if it's a triangle at all, then AC is a straight line, and it goes through the origin. This means it's a diameter, and any inscribed triangle with one side as the diameter is a right triangle. From there you should be able to use Pythagoras. Only don't forget that AC is the hypotenuse.

So: 1^2 + AB^2 = 2^2

1 + AB^2 = 4

AB^2 = 3

AB = root 3

Area = (b*h)/2 = 1 * (root 3)/2 = (root 3)/2

Make sense?

-t

Glad to help!

-t

Tommy's absolutely right, but I just wanted to add - you didn't need to use pythagorean theorem here! As he said, we know that this has to be a right triangle, if one side is the diameter (that side will also always be the hypotenuse of the triangle).

If you have a right triangle in which the hypotenuse is twice the length of one of the sides, it must be a 30-60-90 triangle. If you haven't already, you should definitely memorize the side length relationships in a 30-60-90:



If one side is 1, and the hypotenuse is 2, the other side must be root3. So (1/2)(base*height) = (1/2)(1)(root3) = (root3)/2

Recognizing that this is a 30-60-90 might have saved you some time on this one.