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

Note that

$A_{CMD}=\dfrac{1}{2}\cdot MC\cdot h=7\ sq. ft.$

Let H be the height of triangle ABC. Since $\dfrac{MD}{DB}=\dfrac{1}{2},$ then

$\dfrac{H}{h}=\dfrac{5}{1}, \\ \\H=5h.$

1.

$A_{BDC}=A_{MBC}-A_{CMD}=\dfrac{1}{2}\cdot MC\cdot H-\dfrac{1}{2}\cdot MC\cdot h=\dfrac{1}{2}\cdot MC\cdot (5h-h)=\\ \\=4\cdot \dfrac{1}{2}\cdot MC\cdot h=4\cdot 7=28 sq. ft.$

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

$A_{AMB}=\dfrac{1}{2}\cdot AM\cdot H=\dfrac{1}{2}\cdot MC\cdot 5h=5\cdot \dfrac{1}{2}\cdot MC\cdot h=5\cdot 7=35\ sq. ft.$

3.

$A_{ABC}=\dfrac{1}{2}\cdot AC\cdot H=\dfrac{1}{2}\cdot 2MC\cdot 5h=10\cdot \dfrac{1}{2}\cdot MC\cdot h=10\cdot 7=70\ sq. ft.$

Answer:

$A_{BDC}=28\ sq. ft,\ A_{AMB}=35\ sq. ft,\ A_{ABC}=70\ sq. ft.$