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umka21 [38]
2 years ago
15

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.

Mathematics
2 answers:
ddd [48]2 years ago
5 0

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

OLga [1]2 years ago
3 0

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, <em>A</em>BMP=<em>A</em>CMB=21m^2 which is due to transitivity.
  2. <em>A</em>PCB=<em>A</em>BMP+<em>A</em>CMB=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: <em>A</em>ACP=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 <em>A</em>ABC sooner.) <em>A</em>ABC=<em>A</em>ACP+<em>A</em>PCB=14+42=56m^2
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