6
2 rhombuses and 2 triangles
4 sqaure
4
Triangles are needed to form polygons of 3 sides or more as for instance a 6 sided hexagon needs 4 triangles and a 10 sided decagon needs 8 triangles.
Yes, CPCTC (Corresponding Parts of Congruent Triangles are Congruent) can be used multiple times in a proof, provided that you have established the congruence of the triangles involved. Each instance of CPCTC can be applied to demonstrate the congruence of different corresponding parts as needed throughout the proof. Just ensure that the triangles being referenced are congruent before applying CPCTC.
To create an irregular hexagon using 6 equal triangles, start by arranging the triangles so that their bases align to form the perimeter of the hexagon. Position three triangles in one row with their bases touching and the apexes pointing outward, then place the remaining three triangles on top, ensuring their bases align with the points where the first three triangles meet. Adjust the angles of the triangles as needed to achieve an irregular shape, ensuring they still connect at their vertices. Finally, check that all triangles are equal and the overall shape maintains the hexagonal structure.
The AA similarity postulate states that if two angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar. However, the AA congruence postulate is not needed because knowing two angles of one triangle are congruent to two angles of another triangle doesn't guarantee that the triangles are congruent, as the side lengths can still be different.
Without seeing the picture, I can't tell what's already known to be congruent, so there's no way I can figure out what 'else' is needed.
never tried it without my Gay lines so I wouldn't know for sure, but try 3 and see what you get
To make six sets using rhombuses, trapezoids, triangles, and hexagons, you would need a total of six sets. Each set would consist of one of each shape - a rhombus, a trapezoid, a triangle, and a hexagon. Therefore, you would need 6 sets x 4 shapes per set = 24 shapes in total to make six sets using these shapes.
6
To prove triangle TUX is congruent to triangle DEO by the Hypotenuse-Leg (HL) theorem, we need to establish that both triangles are right triangles. Specifically, we need to confirm that the hypotenuse of triangle TUX is equal to the hypotenuse of triangle DEO, and that one leg of triangle TUX is equal to one leg of triangle DEO. Additionally, we should identify the right angles in both triangles to validate their classification as right triangles.