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.
[tex]\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}[/tex]
Since the area of BPC is [tex]\frac{3}{4}[/tex]th part of total area, therefore area of APC is [tex]\frac{1}{4}[/tex]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.
[tex]\text{Area of } \triangle BMP=\text{Area of } \triangle BMC[/tex]
[tex]21=\text{Area of } \triangle BMC[/tex]
Area of BPC is,
[tex]\text{Area of } \triangle BPC=\text{Area of } \triangle BMP+\text{Area of } \triangle BMC[/tex]
[tex]\text{Area of } \triangle BPC=21+21[/tex]
[tex]\text{Area of } \triangle BPC=42[/tex]
Area of APC is,
[tex]\text{Area of } \triangle APC=\frac{1}{3}\times \text{Area of } \triangle BPC[/tex]
[tex]\text{Area of } \triangle APC=\frac{1}{3}\times 42[/tex]
[tex]\text{Area of } \triangle APC=14[/tex]
Area of ABC is,
[tex]\text{Area of } \triangle ABC=\text{Area of } \triangle APC+\text{Area of } \triangle BPC[/tex]
[tex]\text{Area of } \triangle ABC=14+42=56[/tex]
Therefore, the area of ABC is 56 m².
