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
Solution: 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
