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faust18 [17]
3 years ago
11

n △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 21m^2.

Mathematics
1 answer:
monitta3 years ago
5 0

Answer: The area of ABC is 56 m².

Explanation:

It is given that in △ABC, point P∈ AB is so that AP:BP=1:3 and point M is the midpoint of segment CP.

Since point P divides the line AB in 1:3, therefore the area of triangle APC and BPC is also in ratio 1:3. To prove this draw a perpendicular h on AB from C.

\frac{\text{Area of } \triangle BCP}{\text{Area of } \triangle ABC} =\frac{\frac{1}{2}\times BP\times CH}{\frac{1}{2}\times AB\times CH} =\frac{BP}{AB}= \frac{3}{4}

Since the area of BPC is \frac{3}{4}th part of total area, therefore area of APC is  \frac{1}{4}th part of total area.

The point M is the midpoint of CP, therefore the area of BMP and BMC is equal by midpoint theorem.

\text{Area of } \triangle BMP=\text{Area of } \triangle BMC

21=\text{Area of } \triangle BMC

Area of BPC is,

\text{Area of } \triangle BPC=\text{Area of } \triangle BMP+\text{Area of } \triangle BMC

\text{Area of } \triangle BPC=21+21

\text{Area of } \triangle BPC=42

Area of APC is,

\text{Area of } \triangle APC=\frac{1}{3}\times \text{Area of } \triangle BPC

\text{Area of } \triangle APC=\frac{1}{3}\times 42

\text{Area of } \triangle APC=14

Area of ABC is,

\text{Area of } \triangle ABC=\text{Area of } \triangle APC+\text{Area of } \triangle BPC

\text{Area of } \triangle ABC=14+42=56

Therefore, the area of ABC is 56 m².

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