Question

In trapezoid ABCD with legs AB and CD , diagonals BD ∩ AC =M so that BM:MD=1:4. Find AAMD, ACMD, and AACD if the area of AABM=8 in2.

Answer 1

If BM : MD = 1 : 4, then
`BM=x\ in,\ MD=4x\ in.` Triangles AMD and CMB are similar, then

`(BM)/(MD)=(CM)/(MA)=(1)/(4).`

Let
`CM=y\ in,` then
`MA=4y\ in.`

Given
`A_(\triangle ABM)=8\ in^2.` Note that

`A_(\triangle ABM)=(1)/(2)\cdot BM\cdot MA\cdot \sin\alpha=(1)/(2)\cdot x\cdot 4y\cdot \sin\alpha=8,\\ \\xy\sin\alpha=4.`

1. Consider triangle CMD. The area of this triangle is

`A_(\triangle CMD)=(1)/(2)CM\cdot MD\cdot \sin\alpha=(1)/(2)\cdot y\cdot 4x\cdot \sin\alpha=2xy\sin\alpha=2\cdot 4=8\ in^2.`

2. Consider triangle AMD. The area of this triangle is

`A_(\triangle AMD)=(1)/(2)\cdot AM\cdot MD\cdot \sin(180^(\circ)-\alpha)=(1)/(2)\cdot 4y\cdot 4x\cdot \sin\alpha=8xy\sin\alpha=8\cdot 4=32\ in^2.`

3. Consider triangle ACD. The area of this triangle is

`A_(\triangle ACD)=A_(\triangle AMD)+A_(\triangle CMD)=32+8=40\ in^2.`

Answer:
`A_(\triangle CMD)=8\ in^2,\ A_(\triangle AMD)=32\ in^2,\ A_(\triangle ACD)=40\ in^2.`