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

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

Answer 2

Answer:

Area of $\rm \bigtriangleup ABC = 56 m^2$

Step-by-step explanation:

Given :

In $\rm \bigtriangleup ABC$ ,

$\rm \dfrac {AP}{BP} = \dfrac {1}{3}$

M is the midpoint of segment CP.

Calculation :

BMP and CMB are the triangles whose area are same because there height and base are same. And hence area of  $\rm \bigtriangleup BMP = \rm \bigtriangleup CMB = 21^2$ which is due to transitivity.

Area of triangle PCB is equal to the combine area of triangles BMP and CMB which is equal to $\rm 42 m^2$.

Now, area of triangle ACP

$=42\times \dfrac {1}{3}= 14$

Now we add all the areas, we get the are of triangle ABC

$\rm = 14+42=56m^2$

Therefore, area of $\rm \bigtriangleup ABC = 56 m^2$

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