A triangle with area 64 is inscribed in a semi-circle. What is the area of the semi-circle?
(1) The largest triangle that can be inscribed in the semi-circle has an area of 64.
(2) The diameter of the semi-circle is also one side of the triangle
Triangle 1 - DS
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- karthikpandian19
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The largest triangle that can be inscribed would be a 45-45-90 triangle with diameter as hypotenuse.
1) since its a 45-45-90 triangle b = h
1/2 * b^2 = 64
b^2 = 128
b = h ~ 11.3
Therefore diameter (hypotenuse) will be Sqrt(2)*11.3 - Area can be found using radius. Sufficient.
(2) Given diameter is one side of triangle, multiple triangles (with different areas) can be drawn. Given area = 64, we cannot determine other 2 angles. (Could be 30-60-90 or 45-45-90 or anything else)
A IMO
1) since its a 45-45-90 triangle b = h
1/2 * b^2 = 64
b^2 = 128
b = h ~ 11.3
Therefore diameter (hypotenuse) will be Sqrt(2)*11.3 - Area can be found using radius. Sufficient.
(2) Given diameter is one side of triangle, multiple triangles (with different areas) can be drawn. Given area = 64, we cannot determine other 2 angles. (Could be 30-60-90 or 45-45-90 or anything else)
A IMO
- karthikpandian19
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- chufus
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here is a shorter approach. You correctly identified that the largest triangle will be an isoceles one. In that case, height would be half the diameter, that is the radius. If radius is x, diameter is 2x
Let base is diameter = 2x
Let Height is radius = x
so (1/2)*2x^2 = 64
solve for x = 8
So radius is 8. Hence A is sufficient !
Let base is diameter = 2x
Let Height is radius = x
so (1/2)*2x^2 = 64
solve for x = 8
So radius is 8. Hence A is sufficient !