Question

In △ABC, point M is the midpoint of AC , point D∈ BM so that MD:DB=1:4. If ACMD=7 ft2, find ABDC, AAMB, and AABC.

Answer 1

28,35,70

Explanation:

Answer 2

Note that

`A_(CMD)=(1)/(2)\cdot MC\cdot h=7\ sq. ft.`

Let H be the height of triangle ABC. Since
`(MD)/(DB)=(1)/(2),` then

`(H)/(h)=(5)/(1), \\ \\H=5h.`

1.

`A_(BDC)=A_(MBC)-A_(CMD)=(1)/(2)\cdot MC\cdot H-(1)/(2)\cdot MC\cdot h=(1)/(2)\cdot MC\cdot (5h-h)=\\ \\=4\cdot (1)/(2)\cdot MC\cdot h=4\cdot 7=28 sq. ft.`

2. M is midpoint of AC, then AM=MC.

`A_(AMB)=(1)/(2)\cdot AM\cdot H=(1)/(2)\cdot MC\cdot 5h=5\cdot (1)/(2)\cdot MC\cdot h=5\cdot 7=35\ sq. ft.`

3.

`A_(ABC)=(1)/(2)\cdot AC\cdot H=(1)/(2)\cdot 2MC\cdot 5h=10\cdot (1)/(2)\cdot MC\cdot h=10\cdot 7=70\ sq. ft.`

Answer:

`A_(BDC)=28\ sq. ft,\ A_(AMB)=35\ sq. ft,\ A_(ABC)=70\ sq. ft.`