Question

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.

Answer 1

M is mid point of CP. M will divide the $\Delta$ BPC in two equal parts $\Delta$ BMC and    $\Delta$ BMP.

Area of $\Delta$ BMP is equal to 21m^2

Since, $\Delta$ BMC = $\Delta$ BMP

Area of  $\Delta$ BPC = Area of  $\Delta$ BMC +Area of $\Delta$ BMP =  21 + 21 = 42m^2

and since ratio of AP:BP =1:3  so the area of $\Delta$ BMP will be 1/3 of Area of $\Delta$ ABC

hence, Area of $\Delta$ ABC = 63m^2

Answer 2

Answer: 56 m^2

Step-by-step explanation:

1. 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.
2. APCB=ABMP+ACMB=21+21=42m^2
3. The 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.
4. After that, we use Substitution and Algebra to calculate the area of: AACP=$42 / 3 * 1 = 14 * 1 = 14$
5. 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