In trapezoid abcd with legs ab and cd , diagonals bd ∩ ac =m so that bm: md=1: 4. find area of amd, cmd, and acd if the area of aabm=8 in2.

Point M divides both diagonals in the ratio 1:4. Using AAA, you can show that ∆BCM ~ ∆DAM. Since corresponding parts of similar triangles have the same ratio, we know CM:MA = BM:MD = 1:4.

The height from A to line BD is the same for triangles ABM and AMD. Since MD = 4 × BM, we must have AAMD = 4 × AABM = 32 in².

Likewise, the height from D to line AC is the same for both triangles AMD and CMD. Since MC = 1/4 × MA, the area ACMD = 1/4 × AAMD = 8 in².

The area of triangle ACD is the sum of the areas of triangles AMD and CMD, so is ...

... AACD = AAMD + ACMD = 32 in² +8 in² = 40 in²

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In the attached figure, the black horizontal line at the top is parallel to BD, so point A can be moved anywhere along that line and the areas of triangles ABM and AMD remain constant. The figure geometry changes, but the remaining areas also remain constant (as the above arguments show). It can be satisfying to have a little toy to play with to confirm the math. This one is provided by GeoGebra.

idk

step-by-step explanation: