In △abc, point p∈ ab is so that ap: bp=1: 3 and point m is the midpoint of segment cp . find the area of △abc if the area of △bmp is equal to 21m2.

56 m^2

Step-by-step explanation:

The areas of triangle BMP and CMB are equal because the base and height are the same. Therefore, ABMP=ACMB=21m^2 which is due to transitivity.APCB=ABMP+ACMB=21+21=42m^2The ratio of the areas of triangles ACP to PCB is the same as their bases AP and BP which is 1:3. This is due to area of triangles.After that, we use Substitution and Algebra to calculate the area of: AACP=Finally, we add all of the areas together. (Or you could've just multiplied in the last equation by 4 rather than 3 to get the AABC sooner.) AABC=AACP+APCB=14+42=56m^2

Use a graph site then you will find your answer