AD Median is drawn to side BC which is the vertex of ∆ABC.
Median AD divides AABC into two triangles ∆ABD and ∆ADC which are equal in area.
Now Median AD bisects BC.
∴ BD = DC.
(i) As per Mid-Points formula,
Coordinates of D are
= \(\sqrt{(4)^2 - (0)^2}\)
= \(\sqrt{16, 0}\)
= (4,0)
∴ Coordinates of D are (4, 0)
(ii) Now, Area of ∆ABD:
(iiii) Now, Area of ADC:
= – 3 sq. units.
∴ Area of ∆ADC = -3 sq. units.
∴ Area of ∆ABC : = Area of ∆ABD + Area of ∆ADC
= (-3) +(-3)
= -6 sq. units.
(iv) Now, Area of ∆ABC : (Direct Method)