Question

in trapezoid abcd with legs \\( \overline{ab} \\) and \\( \overline{cd} \\), diagonals \\( \overline{bd} \cap \overline{ac} = m \\) such that \\( bm:md = 1:4 \\). find \\( a_{amd} \\), \\( a_{cmd} \\), and \\( a_{acd} \\) if the area of \\( a_{abm} = 8 \text{ in}^2 \\). answer: \\( a_{adm}=32 \text{ in}^2 \\), \\( a_{cmd}= \\) in.\\( ^2 \\), \\( a_{acd}= \\) in.\\( ^2 \\)

Explanation:

Step1: Find the area of $\triangle AMD$

Since $\triangle ABM$ and $\triangle AMD$ share the same height (the perpendicular distance between the parallel sides of the trapezoid), and the ratio of their bases $BM:MD = 1:4$.
The formula for the area of a triangle is $A=\frac{1}{2}bh$. For two triangles with the same height $h$, the ratio of their areas is equal to the ratio of their bases.
Let $A_{ABM}=\frac{1}{2}BM\times h = 8$ and $A_{AMD}=\frac{1}{2}MD\times h$.
Since $\frac{BM}{MD}=\frac{1}{4}$, then $\frac{A_{ABM}}{A_{AMD}}=\frac{BM}{MD}$.
Substituting $A_{ABM} = 8$ and $\frac{BM}{MD}=\frac{1}{4}$, we get $A_{AMD}=4\times A_{ABM}$.
So $A_{AMD}=4\times8 = 32$ in.$^{2}$.

Step2: Prove $\triangle ABM\sim\triangle CDM$

Because $AB\parallel CD$ in trapezoid $ABCD$, then $\angle ABM=\angle CDM$ and $\angle BAM=\angle DCM$ (alternate - interior angles).
By the AA (angle - angle) similarity criterion, $\triangle ABM\sim\triangle CDM$.
The ratio of the sides $BM:MD = 1:4$, so the ratio of the areas of similar triangles $\triangle ABM$ and $\triangle CDM$ is $A_{ABM}:A_{CDM}=BM^{2}:MD^{2}$.
Since $BM:MD = 1:4$, then $A_{ABM}:A_{CDM}=1:16$.
Given $A_{ABM} = 8$ in.$^{2}$, so $A_{CDM}=16\times8=128$ in.$^{2}$.

Step3: Calculate the area of $\triangle ACD$

The area of $\triangle ACD=A_{AMD}+A_{CMD}$.
We know $A_{AMD} = 32$ in.$^{2}$ and $A_{CMD}=128$ in.$^{2}$.
So $A_{ACD}=32 + 128=160$ in.$^{2}$.

Answer:

$A_{ADM}=32$ in.$^{2}$, $A_{CMD}=128$ in.$^{2}$, $A_{ACD}=160$ in.$^{2}$