Solution:
Given:
D is the mid-point of BC, DE⊥AC
AC = 12 cm, DE = 5 cm
To find:
Area of triangle ΔABC
Here,
- Area of triangle ΔADC = \(\frac12\times AC\times DE\) [∵ Area of triangle is half the product of base and height ]
= \(\frac12\times12\times5\)
= 30 cm2
- Area of triangle ΔABC = 2 x Area if triangle ΔADC [∵ median AD bisects ΔABC into two equal halves by area ]
Or, Area of triangle ΔABC = 2 x 30 cm2 = 60 cm2
Hence the required area of triangle ΔABC is 60 cm2