Question

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.

Answer 1

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.