Montana Mick
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- Jul 21, 2012
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A triangle has sides 104, 112 and 120 cms in length. Find a point that lies equidistant from each corner ? I can find the area, not sure how to proceed. Thanks in advance.
A triangle has sides 104, 112 and 120 cms in length. Find a point that lies equidistant from each corner ? I can find the area, not sure how to proceed. Thanks in advance.
A triangle has sides 104, 112 and 120 cms in length.
Find a point that lies equidistant from each vertex.
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0 112
A triangle has sides 104, 112 and 120 cms in length. Find a point that lies equidistant from each corner ? I can find the area, not sure how to proceed. Thanks in advance.
hello, montana mick!
we want the circumcenter, the center of the circumscribed circle.
It is the intersection of the perpendicular bisectors of the sides.
Perhaps a diagram will help.
Code:| | (40,96) | o | *: * | * : * | * : * | * : * |* : * - - o - - + - - - - - o - - - 0 112
hint:
In a circle, the center lies equidistant from all the points on the circle.
The end-points of a given line segment lies equidistant from any point on a perpendicular bisector of the given line segment.
please share your work with us, indicating exactly where you are stuck - so that we may know where to begin to help you.
A triangle has sides 104, 112 and 120 cms in length. Find a point that lies equidistant from each corner ? I can find the area, not sure how to proceed. Thanks in advance.
i got an answer of 16 , used formula a/2p , a = area, p = perimeter, answer given is 65
A triangle has sides 104, 112 and 120 cms in length. Find a point that lies equidistant from each corner ? I can find the area, not sure how to proceed. Thanks in advance.
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A D C
Familiarity with Heronian triangles yields AB = 13, BC = 14, CA = 15, BD = 12, AD = 5 and CD = 10 after dividing all sides by 8.
The point you seek is the intersection of the three altitudes, the center of the circumcircle.-------> NO
By means of the Law of Sines and Cosines, you can now determine the angle(s) and lengths of all segments that define the circumcircle center.
Remember to multiply your answers by 8 as we divided the original triangle by 8 to get the Heronian triangle to work with.
The intersection of three altitudes is called the orthocenter. The circumcenter, which is the intersection of the perpendicular bisector of the sides, is what is needed. That point is the center of a circle circumscribed about the triangle and thus the distance from the circumcenter to each vertex of the triangle is the radius of the circle and thus are equal.
A triangle has sides 104, 112 and 120 cms in length. Find a point that lies equidistant from each corner ? I can find the area, not sure how to proceed. Thanks in advance.