Note that
[tex]A_{CMD}=\dfrac{1}{2}\cdot MC\cdot h=7\ sq. ft.[/tex]
Let H be the height of triangle ABC. Since [tex]\dfrac{MD}{DB}=\dfrac{1}{2},[/tex] then
[tex]\dfrac{H}{h}=\dfrac{5}{1}, \\ \\H=5h.[/tex]
1.
[tex]A_{BDC}=A_{MBC}-A_{CMD}=\dfrac{1}{2}\cdot MC\cdot H-\dfrac{1}{2}\cdot MC\cdot h=\dfrac{1}{2}\cdot MC\cdot (5h-h)=\\ \\=4\cdot \dfrac{1}{2}\cdot MC\cdot h=4\cdot 7=28 sq. ft.[/tex]
2. M is midpoint of AC, then AM=MC.
[tex]A_{AMB}=\dfrac{1}{2}\cdot AM\cdot H=\dfrac{1}{2}\cdot MC\cdot 5h=5\cdot \dfrac{1}{2}\cdot MC\cdot h=5\cdot 7=35\ sq. ft.[/tex]
3.
[tex]A_{ABC}=\dfrac{1}{2}\cdot AC\cdot H=\dfrac{1}{2}\cdot 2MC\cdot 5h=10\cdot \dfrac{1}{2}\cdot MC\cdot h=10\cdot 7=70\ sq. ft.[/tex]
Answer:
[tex]A_{BDC}=28\ sq. ft,\ A_{AMB}=35\ sq. ft,\ A_{ABC}=70\ sq. ft.[/tex]
