ISOSCELES TRIANGLE
- Thread starter hope
- Start date
- Joined
- Apr 12, 2005
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- 11,339
First, you may wish to get the right drawing. In your post on the other site, you have it rotated 90°.
http://img309.imageshack.us/img309/5/in ... ola7mv.png
1) Draw the parabola
2) Inscribe a triangle (Ignore that it should be a right triangle, for now.)
2a) Move out a bit on the x-axis and draw a vertical line.
2b) It intersects the parabola in two places
2c) Draw a line segment connecting these intersections with the origin.
3) Label the points of intersection in some general way.
3a) The one in the first quadrant, above the x-axis, is (x<sub>0</sub>,y<sub>0</sub>) = (¼*(y<sub>0</sub>)<sup>2</sup>,y<sub>0</sub>)
3b) The one below the x-axis is, by symmetry, (¼*(y<sub>0</sub>)<sup>2</sup>,-y<sub>0</sub>)
4) Using the vertical line segment as the base, the area of the triangle is ½*(2*y<sub>0</sub>)*x<sub>0</sub> = ½*(2*y<sub>0</sub>)*¼*(y<sub>0</sub>)<sup>2</sup> = ¼*(y<sub>0</sub>)<sup>3</sup>. You may have to ponder this for a while. You must know where it comes from.
5) Set that equal to 16 and solve for y<sub>0</sub> and subsequently for x<sub>0</sub>
6) It is only by luck that this is a Right triangle. That restriction may have made the problem without solution. In this case, it didn't cause us any problems.
http://img309.imageshack.us/img309/5/in ... ola7mv.png
1) Draw the parabola
2) Inscribe a triangle (Ignore that it should be a right triangle, for now.)
2a) Move out a bit on the x-axis and draw a vertical line.
2b) It intersects the parabola in two places
2c) Draw a line segment connecting these intersections with the origin.
3) Label the points of intersection in some general way.
3a) The one in the first quadrant, above the x-axis, is (x<sub>0</sub>,y<sub>0</sub>) = (¼*(y<sub>0</sub>)<sup>2</sup>,y<sub>0</sub>)
3b) The one below the x-axis is, by symmetry, (¼*(y<sub>0</sub>)<sup>2</sup>,-y<sub>0</sub>)
4) Using the vertical line segment as the base, the area of the triangle is ½*(2*y<sub>0</sub>)*x<sub>0</sub> = ½*(2*y<sub>0</sub>)*¼*(y<sub>0</sub>)<sup>2</sup> = ¼*(y<sub>0</sub>)<sup>3</sup>. You may have to ponder this for a while. You must know where it comes from.
5) Set that equal to 16 and solve for y<sub>0</sub> and subsequently for x<sub>0</sub>
6) It is only by luck that this is a Right triangle. That restriction may have made the problem without solution. In this case, it didn't cause us any problems.