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
Inscribed triangle
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Tommy Wallach
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Tommy Wallach
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Glad to help!
-t
-t
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ceilidh.erickson
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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.
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.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
EdM in Mind, Brain, and Education
Harvard Graduate School of Education
