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

Since M is the midpoint of segment CP, BM is the median of the triangle PBC.

Note that median of a triangle divides it into two triangles of equal area.

Therefore, area (BCP) = 2 × area (BMP)

Given that area (BMP) = 21
`m^(2)`

So, area (BCP) = 2 × 21 = 42
`m^(2)` --- (1)

Let h be the height of triangle ABC from the vertex C.

Then, area of Δ ABC =
`(1)/(2)(AB)(h)`

Area of Δ BCP =
`(1)/(2)(BP)(h)`

Also, since AP : BP = 1 : 3,
`BP = (3)/(4) AB`

So, area of Δ BCP =
`(1)/(2) (3)/(4) AB(h)`

But, from (1) area (BCP) = 42

Therefore,
`(1)/(2) (3)/(4) AB(h)=42`

`(1)/(2) AB(h)=42((4)/(3) )`

`= 56 m^(2)`

Hence, area of Δ ABC =
`= 56 m^(2)`.

Answer 2

Answer: Area of ΔABC = 56m².

Step-by-step explanation:

Since we have given that

point P∈ AB, such that AP:BP=1:3

Point M is the midpoint of segment CP.

Now, we have given that

Area of ΔBMP=21 m²

Since M is the mid-point of CP, so,

Area of ΔCMP= 21 m².

So, Area of ΔBCP = 21+21=42 m²

According to question,

`(3)/(4)* x= 42\\\\x=(42)/(3)=14\\\\\text{ Area of }\triangle APC=14m^2\\\\`

So, area of ΔABC is given by

`42+14=56m^2`

Hence, area of ΔABC =56m².