Final answer:
The areas of triangles APC and PMC in a triangle with a median and an intersecting line is 35m^2.
Step-by-step explanation:
Given: AM is a median in ΔABC where M∈BC.A line drawn through point M intersects AB at its midpoint p.
Also, area of ΔAPM = 35 m².
To Find: Area of △APC and △PMC
In ΔAMB, PM is median.
As you must keep in mind, Median of triangle divides it into triangles of equal area.
→Area(ΔAPM) = Area(BPM)
= 35 m²
So, Area ( ΔABM)= Area(ΔAPM) + Area(BPM)
= 35+35
=70 m²
Now, Area (ΔABM)
= Area(ΔAMC)
=70 m²→[Median of triangle divides it into triangles of equal area.]
→Area (ΔABC)= Area (ΔABM) + Area(ΔAMC)
=70+70
=140 m²
Area(Trpzm BPMC) = Area(ΔABC) - Area(ΔBPM)
= 140 - 35
= 105 m²
PM ║AC, Line joining mid point of two sides of triangle is parallel to third side.
Area(ΔAMC) =Area(ΔPCA)
= 70 m² [Triangle on the same base and between the same parallels are equal in area.]
Area(ΔPMC)= Area(trpzmPMAC) - Area(ΔPAC)
=135 - 70
=35 m²
Hence, the area is 35 m².