Triangle centre

[Montana Mick]

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.

 

[Deleted member 4993]

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.

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.

 

[soroban]

Hello, Montana Mick!

A triangle has sides 104, 112 and 120 cms in length.
Find a point that lies equidistant from each vertex.

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.

        |
        |  (40,96)
        |     o
        |    *: *
        |   * :   *
        |  *  :     *
        | *   :       *
        |*    :         *
    - - o - - + - - - - - o - - -
        0                112

 

[pappus]

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 don't understand your question(?):

1. Are you asked to find this point by construction with compass and unmarked straight edge?

2. Are you asked to determine the coordinates of the point then we need the coordinates of the vertices too.

 

[Montana Mick]

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.

        |
        |  (40,96)
        |     o
        |    *: *
        |   * :   *
        |  *  :     *
        | *   :       *
        |*    :         *
    - - o - - + - - - - - o - - -
        0                112

used the formula p/2a, it worked out as 16 exactly.the answer given is 65! P=perimeter, a = area.

 

[Montana Mick]

Triangle CENTER

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.

i got an answer of 16 , used formula a/2p , a = area, p = perimeter, answer given is 65 !

 

[Deleted member 4993]

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.

You are asked to find the in-center (a point) - and you say

i got an answer of 16 , used formula a/2p , a = area, p = perimeter, answer given is 65

How can a point be 65 - and what formula is that?

If you need to find the area of the triangle (not "a point that lies equidistant from each corner") - first you need to post the correct question, then use Heron's formula.

 

[TchrWill]

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.

B
*
* *
E * * *
* * *
* * * *
* * * *
* ***************** *
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.


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.

 

[srmichael]

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.

B
*
* *
E * * *
* * *
* * * *
* * * *
* ***************** *
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.

 

[TchrWill]

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.

I was well aware of this and cannot figure out, for the life of me, why I typed in "the three altitudes".
Of course, the three perpendicular bisectors of the three sides meet at the center of the circumcircle, the

.........................B
........................*
.....................*.... *
..............E *... *...... *
............* ........*........... *
........* ............* ...............*
....*................ *................... *
* ***************** *****
A.................. D....................... C

AB = 13, BC = 14, CA = 15, BD = 12, AD = 5 and CD = 10 after dividing all sides by 8.

Also, remember that the radius of the circle derives from A = abc/4R where A = the area of the triangle, a, b and c are the three sides of the triangle and R = the radius of the circumcircle.

Thanks for the heads up.

 

[TchrWill]

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.

Not having seen an answer to your problem or whether you have solved it on your own, I offer the following as one solution. A second solution is available if you so desire.

The point you seek is the intersection of the three perpendicular bisectors of the sides., or the center of the circumcircle passing through each vertex..


A revised figure is:

........................B
.......................*
..................*.......*
..............*..............*
..........*......................*
......*..............................*
A ***********************C


Dividing the sides by 8 yields the familiar Heronian triangle AB = 13, BC = 14 and AC = 15 with one integer altitude of 12. (The altitude of 12 can be derived using Heron’s formula knowing only the 3 sides, A = sqrt[s(s-a)(s-b)(s-c)], where a, b and c are the sides and s = (a+b+c)/2 which yields an area of 84. From A = bh/2, h = 2A/b = 12.)


We will now work with the sides of 104, 112, 120 and altitude 96.

Angle BAC derives from (BC)^2 = (AB)^2 + (AC)^2 – 2(AB)(AC)cos /_BAC from which /_BAC = 59.5164º.

The radius of the circumcircle passing through the three vertices derives from the Second Law of Sines with R = the circumcircle radius = a/2sinA = b/2sinB = c/2sinC = R.


Using /_BAC = 59.5164 and BC = 112, R = 112/2(.86177) = 64.9822.

At the mid point of AC, F, construct a perpendicular bisector.

With radius 64.9822, and centers A and C, swing arcs intersecting at point P on the perpendicular bisector above AC.

Using triangle CFP, with AC = 60 and PC = 64.9822, FP = sqrt(64.9822^2 = 60^2 = 24.9536.

Thus, with refernce to the above figure, the circumcircle center is located 60cm to the right of point C and 24.9536cm above AC.