Aww, man. Proofs? You're on your own.
A=1/2bh The area of a triangle is 1/2bh. If the base of it is a triangle and all 4 of the triangles aren't the same, then you have to find the area of the base triangle and then the three other triangles (which should all have the same area). If all four of the triangles have the same area, then just find the area of one of the triangles and multiply that by four. A triangular pyramid that has four equal triangles is also called a tetrahedron.
No. Although the area of every triangle is equal to half the area with the same base and height, only right angled triangles are half a rectangle.
yes
yes they have same area
If you double (2 times) the perimeter the area will will be 4 times larger. Therefore the area is proportional to the square of the perimeter or the perimeter is proportional to the square root of area. The relationship as shown above applies only to triangles with similar proportions, that is when you scale up or down any triangle of fixed proportions. Other than that requirement, there is no relationship between perimeter and area of any shape of triangle except that it can be stated that the area will be maximum when the sides are of equal length (sides = 1/3 of perimeter).
Only if the two triangles are congruent will they have equal areas. A third fact is required to determine they are congruent (and thus have the same area): 1) The third sides are equal; 2) The angles enclosed between the sides are equal; or 3) The same one of the sides is the hypotenuse of the triangles, which are right angled triangles.
Yes.
Yes, they do.
I believe so, though I am not sure I can prove it.
Any triangles where the base multiplied by the height equal 2, so technically, infinity.
Given:In a triangle ABC in which EF BC To prove that:AE/EB=AF/FC Construction:Draw EX perpendicular AC and FY perpendicular AB Proof:taking the ratios of area of triangle AEF and EBF and second pair of ratio of area of triangle AEF and ECF. We get AE/EB and AF/FC we know that triangle lie b/w sme and same base is equal in area therefore area of EBF I equal to area of ECF therefore AE/EB=AF/FC HENCE PROVED
only for triangles and parallelograms as far as i know
A=1/2bh The area of a triangle is 1/2bh. If the base of it is a triangle and all 4 of the triangles aren't the same, then you have to find the area of the base triangle and then the three other triangles (which should all have the same area). If all four of the triangles have the same area, then just find the area of one of the triangles and multiply that by four. A triangular pyramid that has four equal triangles is also called a tetrahedron.
There is not enough information about the triangles to be able to answer the question.
We know that diagonals of parallelogram bisect each other. Therefore, O is the mid-point of AC and BD. BO is the median in ΔABC. Therefore, it will divide it into two triangles of equal areas. Area (ΔAOB) = Area (ΔBOC) ... (1) In ΔBCD, CO is the median. Area (ΔBOC) = Area (ΔCOD) ... (2) Similarly, Area (ΔCOD) = Area (ΔAOD) ... (3) From equations (1), (2), and (3), we obtain Area (ΔAOB) = Area (ΔBOC) = Area (ΔCOD) = Area (ΔAOD) Therefore, it is evident that the diagonals of a parallelogram divide it into four triangles of equal area.
5, 12, 13 and 6, 8, 10
AAS: If Two angles and a side opposite to one of these sides is congruent to thecorresponding angles and corresponding side, then the triangles are congruent.How Do I know? Taking Geometry right now. :)