The formula is A=1/2bh b=base of triangle h=height of triangle A=area 1/2=divide base times height by 2
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Only if the two triangles have the same base and height then they have the same area, because an area of a triangle OS the base times the height divided by two.
the formula for finding the area of an ellipse is add it then multiply and subtract that is the final
There are basically two techniques for finding the area of a shape with uneven or irregularly shaped sides. If the sides can be described by algebraic equations, then integral calculus can be used to find the area. Failing that, you can approximate the irregular shape by fitting in a number of smaller, regularly shaped polygons such as squares and triangles, whose area can be calculated by simple geometric techniques.
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.
they are all postulates or shortcuts on finding 2 triangles congruence, except that SAA does not exist.