First, draw your diagram. Make sure all of the specifications are there.
AD : DB = 2:3
Area of ΔABC = 30 in²
This means that the sides that make up AB (AD and DB) are 2:3. The total length of side AB = 5. The altitude of ΔABC is CD, and that is 0.5 (5 × x), and if you solve that, x = 12. So CD = 12.
Since point M is the midpoint, it divides CD into 2 equal lengths, DM and CM, both of which equal 6 (12 ÷ 2 = 6)
Now:
ΔDMB = 0.5 (6 × 3) = 0.5 (18) = 9 in²
ΔDMA = 0.5 (6 × 2) = 0.5 (12) = 6 in²
ΔACD = 5/2 = 30/x, 5x = 60, x = 12 in² (Main Problem 2: If 2 triangles have congruent altitudes, then the ratio of their areas is equal to the ratio of the sides to which those altitudes are drawn.)
ΔAMC = 12 in² - 6 in² = 6 in² (Area of ΔACD - Area of ΔDMA = Area of ΔAMC)
ΔBDC = 5/3 = 30/x, 5x = 90, x = 18 in² (Main Problem 2: If 2 triangles have congruent altitudes, then the ratio of their areas is equal to the ratio of the sides to which those altitudes are drawn.)
ΔBMC = 18 in² - 9 in² = 9 in² (Area of ΔBDC - Area of ΔDMA = Area of ΔBMC)
So the answers are:
Area of ΔACD = 12 in²
Area of ΔAMC = 6 in²
Area of ΔDMB = 9 in²
Area of ΔCMB= 9 in²
Hope this helped!
