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∙ 12y agotwo slightly smaller triangles of equal size :)
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∙ 12y agoThe medians of a triangle are concurrent and the point of concurrence, the centroid, is one-third of the distance from the opposite side to the vertex along the median
Draw a line from any vertex to a point on one of the adjacent sides and cut along it.
The area of a triangle is one-half the product of the triangle's base and height. The height of an equilateral triangle is the distance from one vertex along the perpendicular bisector line of the opposite side. This line divides the equilateral triangle into two right triangles, each with a hypotenuse of 9c and a base of (9/2)c. From the Pythagorean theorem, the height must be the square root of {(9c)2 - [(9/2)c]}, and this height is the same as that of the equilateral triangle.
Because they represent a pair of coordinates
Start with an equilateral triangle with all sides of length 2 units and all angles of 60 degrees. Draw the altitude from the apex (top vertex) to the base. Since this is an equilateral triangle, it is easy to show that this line bisects the base.So now you have a right angled triangle, with a base of 1 and a hypotenuse of 2. Therefore, by Pythagoras, its vertical height is sqrt(3).Then tan(60) = opposite side/adjacent side = sqrt(3)/1 = 1.7321It is probably easier if you sketch the diagrams as you go along.
A 30-60-90 right triangle
The medians of a triangle are concurrent and the point of concurrence, the centroid, is one-third of the distance from the opposite side to the vertex along the median
Draw a line from any vertex to a point on one of the adjacent sides and cut along it.
Suppose you have a triangle whose vertices are A, B and C, and the sides opposite these vertices are a, b and c, respectively. Cut out the triangle. Fold it so that vertex B meets vertex C. Mark the point where this fold is on side a. Mark this point as D and fold along AD. Fold it so that vertex A meets vertex C. Mark the point where this fold is on side b. Mark this point as E and fold along BE. Fold it so that vertex A meets vertex B. Mark the point where this fold is on side c. Mark this point as F and fold along CF. The three folds, AD, BE and CF meet at the circumcentre. You do not need all three - any two of them will do.
Infinitely many. Take two triangles. Let a vertex of triangle A just touch the side of the other triangle, B. Change the angle of triangle A. there are infinitely many angles and so infinitely many orientations for A.Not only that. Slide triangle A along the side of triangle B. There are infinitely many points on the side of B so that's an infinite number of shapes.And that is with just two triangles!Infinitely many. Take two triangles. Let a vertex of triangle A just touch the side of the other triangle, B. Change the angle of triangle A. there are infinitely many angles and so infinitely many orientations for A.Not only that. Slide triangle A along the side of triangle B. There are infinitely many points on the side of B so that's an infinite number of shapes.And that is with just two triangles!Infinitely many. Take two triangles. Let a vertex of triangle A just touch the side of the other triangle, B. Change the angle of triangle A. there are infinitely many angles and so infinitely many orientations for A.Not only that. Slide triangle A along the side of triangle B. There are infinitely many points on the side of B so that's an infinite number of shapes.And that is with just two triangles!Infinitely many. Take two triangles. Let a vertex of triangle A just touch the side of the other triangle, B. Change the angle of triangle A. there are infinitely many angles and so infinitely many orientations for A.Not only that. Slide triangle A along the side of triangle B. There are infinitely many points on the side of B so that's an infinite number of shapes.And that is with just two triangles!
There is no such thing as an equator triangle. A triangle with its base along the equator and its other vertex elsewhere can have angles adding up to just over 180 to just under 540 degrees. The nearer to the pole, and the longer the base, the greater the angular sum.
The vertex of a prism is a corner along the side of the prism. There are 8 vertices on a rectangular prism.
The area of a triangle is one-half the product of the triangle's base and height. The height of an equilateral triangle is the distance from one vertex along the perpendicular bisector line of the opposite side. This line divides the equilateral triangle into two right triangles, each with a hypotenuse of 9c and a base of (9/2)c. From the Pythagorean theorem, the height must be the square root of {(9c)2 - [(9/2)c]}, and this height is the same as that of the equilateral triangle.
The surface generated by a straight line, the generator, passing through a fixed point, the vertex, and moving along a fixed curve, the directrix.A right circular cone.
The center of mass of a uniform triangular lamina lies at the intersection of the medians of the triangle, which is also known as the centroid. It is located one-third of the distance from each vertex along each median.
The center of gravity of a triangular lamina lies at the point of intersection of the medians of the triangle, which is also known as the centroid. It is located one-third of the distance from each vertex to the midpoint of the opposite side along the median.
It depends on what the motion is. If the square is sliding along a straight line then the path of the vertex is a straight line. If the square is rotating, the answer will vary according to the location of the centre of rotation.