A circle !!!!!!
It could be a right angle triangle
It could be but does not have to be... (Perpendicular to what?)
A traingle covers half the area of a rectangle with the same base and [perpendicular] height.
An equilateral quadrilateral is not a well defined shape - it could be a rhombus with one angle able to take ANY value. In the same way, an equilateral dodecagon is not a well defined shape and so it is impossible to give a sensible answer to the question.
The description seems a bit confusing (to me) but it sounds like it could be a perpendicular bisector of a side of a triangle.
The median is a line from a vertex to the midpoint of the opposite line and an altitude is a line from a vertex to the opposite line which is perpendicular to the line. These are NOT the same thing in most triangles. The only time they could be the same is in an equilateral triangle.
If the compass angle is changed, the entire geometric shape being drawn is different. For example, if a triangle is being drawn, it could change from an obtuse triangle from a ninety degree triangle.
It could be, but without the diagram it is not possible to be certain.
You have points A, B, and C. Using a compass and straight edge, find a perpendicular bisector of AB (that is, a line that is perpendicular to AB and intersects AB at the midpoint of AB. Next, find a perpendicular bisector of BC. The two lines you found will meet at the center of the circle.
In the same way that you bisect an acute triangle. Alternatively, you could extend one of the rays of the obtuse angle so that you have an acute angle. Bisect that angle and then draw a perpendicular to the bisector of the acute angle through the vertex.
The median of an isosceles triangle from its apex is also the perpendicular bisector of the base. This line divides the triangle into two congruent right angled triangles whose hypotenuse is 3 feet and whose apical angle is 35/2 = 17.5 degrees. If the base of the original triangle was 2b cm then sin(17.5) = b/3 so that b = 3*sin(17.5) = 0.9cm so that the base was 2b = 1.8 feet Alternatively, you could use the sine rule on the triangle:
Not necessarily. It is a plane shape which could be horizontal.
A circle !!!!!!
It could be a right angle triangle
The answer depends on what point of concurrency you are referring to. There are four segments you could be talking about in triangles. They intersect in different places in different triangles. Medians--segments from a vertex to the midpoint of the opposite side. In acute, right and obtuse triangles, the point of concurrency of the medians (centroid) is inside the triangle. Altitudes--perpendicular segments from a vertex to a line containing the opposite side. In an acute triangle, the point of concurrency of the altitudes (orthocenter) is inside the triangle, in a right triangle it is on the triangle and in an obtuse triangle it is outside the triangle. Perpendicular bisectors of sides--segments perpendicular to each side of the triangle that bisect each side. In an acute triangle, the point of concurrency of the perpendicular bisectors (circumcenter) is inside the triangle, in a right triangle it is on the triangle and in an obtuse triangle it is outside the triangle. Angle bisectors--segments from a vertex to the opposite side that bisect the angles at the vertices. In acute, right and obtuse triangles, the point of concurrency of the angle bisectors (incenter) is inside the triangle.
It could be but does not have to be... (Perpendicular to what?)