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lapo4ka [179]
3 years ago
6

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:
erma4kov [3.2K]3 years ago
6 0

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}.

matrenka [14]3 years ago
3 0

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.

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