A triangle inscribed in a circle having a diameter as one side is a right triangle.
ABC is a right angled triangle.Angle B is 90.
So AB^2=AC^2-BC^2=4-1=3
-->AB=sqrt(3)
Area of right angled triangle ABC=1/2*(Product of sides containing 90)=1/2*(BC*AB)
=1/2*(sqrt(3)*1)=sqrt(3)/2
gmatprep - geometry 2
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- Gurpinder
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selango wrote:A triangle inscribed in a circle having a diameter as one side is a right triangle.
Ahhhh...
but if its a right triangle and BC =1 and AC = 2, then should the area be 1?
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krazy800
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AC is the hypotenuse here..Gurpinder wrote:selango wrote:A triangle inscribed in a circle having a diameter as one side is a right triangle.
Ahhhh...
but if its a right triangle and BC =1 and AC = 2, then should the area be 1?
so AB = (3)^0.5
Therefore area = 1/2 * AB * BC
= (3)^.5/2
HTH!!
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