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 = $\frac{1}{2}(AB)(h)$

Area of Δ BCP = $\frac{1}{2}(BP)(h)$

Also, since AP : BP = 1 : 3, $BP = \frac{3}{4} AB$

So, area of Δ BCP = $\frac{1}{2} \frac{3}{4} AB(h)$

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

Therefore, $\frac{1}{2} \frac{3}{4} AB(h)=42$

$\frac{1}{2} AB(h)=42(\frac{4}{3} )$

$= 56 m^{2}$

Hence, area of Δ ABC = $= 56 m^{2}$.

Answer 2

Answer: Area of  ΔABC = 56m².

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,

$\frac{3}{4}\times x= 42\\\\x=\frac{42}{3}=14\\\\\text{ Area of }\triangle APC=14m^2$

So, area of ΔABC is given by

$42+14=56m^2$

Hence, area of  ΔABC =56m².