Answer:
Area of triangle AMD is 32 square inches.
Area of triangle CMD is 8 square inches.
Area of triangle ACD is 40 square inches.
Step-by-step explanation:
In trapezoid ABCD with legs AB and CD.
Diagonal BD and AC intersect at M so that BM:MD=1:4
We are given ar(ABM)=8 in²
Ratio of area of two triangle is equal to ratio of their base.
[tex]\dfrac{ar(ABM)}{ar(AMD)}=\dfrac{BM}{MD}[/tex]
[tex]\dfrac{8}{ar(AMD)}=\dfrac{1}{4}[/tex]
[tex]ar(AMD)=32\text{ in}^2[/tex]
Thus, Area of triangle AMD is 32 square inches.
If two triangles lie on same base and same parallel line then their area is equal.
[tex]\therefore ar(ABD)=ar(ACD)[/tex]
[tex] ar(ACD)=ar(ACD)=ar(AMD)+ar(ABM)[/tex]
[tex] ar(ACD)=8+32\Rightarrow 40\text{ in}^2[/tex]
Thus, Area of triangle ACD is 40 square inches.
[tex]ar(CMD)=ar(ACD)-ar(AMD)[/tex]
[tex]ar(CMD)=40-32\Rightarrow 8\text{ in}^2[/tex]
Thus, Area of triangle CMD is 8 square inches.
