Question

AM is a median in △ABC (M∈ BC ). A line drawn through point M intersects AB at its midpoint P. Find areas of △APC and △PMC, if Area of APM=35m^2.

Answer 1

Answer:

The area of APC is 70m². The area of triangle PMC is 35m².

Step-by-step explanation:

Let the area of triangle ABC be x.

It is given that AM is median, it means AM divides the area of triangle in two equal parts.

$\text{Area of }\triangle ACM=\text{Area of }\triangle ABM=\frac{x}{2}$    .....(1)

The point P is the midpoint of AB, therefore the area of APC and BPC are equal.

$\text{Area of }\triangle APC=\text{Area of }\triangle BPC=\frac{x}{2}$          ......(2)

The point P is midpoint of AB therefore the line PM divide the area of triangle ABM in two equal parts. The area of triangle APM and BPM are equal.

$\text{Area of }\triangle APM=\text{Area of }\triangle BPM=\frac{x}{4}$        .....(3)

The area of triangle APM is 35m².

$\text{Area of }\triangle APM=\frac{x}{4}$

$35=\frac{x}{4}$

$x=140$

Therefore the area of triangle ABC is 140m².

Using equation (2).

$\text{Area of }\triangle APC=\frac{x}{2}$

$\text{Area of }\triangle APC=\frac{140}{2}$

$\text{Area of }\triangle APC=70$

Therefore the area of triangle APC is 70m².

Using equation (3), we can say that the area of triangle BPM is 35m² and by using equation (2), we can say that the area of triangle BPC is 70m².

$\triangle BPC=\triangle BPM+\triangle PMC$

$70=35+\triangle PMC$

$35=\triangle PMC$

Therefore the area of triangle PMC is 35m².