In trapezoid ABCD with bases AD and BC, the diagonals meet at point M. a) Prove that triangles BMC
| April 2, 2021 | education
In trapezoid ABCD with bases AD and BC, the diagonals meet at point M. a) Prove that triangles BMC and DMA are similar. b) Find the area of the triangle DMA if AM: MC = 3: 2, and the area of the triangle BMC is 8cm ^ 2.
In triangles BMC and DMA, the angles BMC and MD are equal as vertical angles, the angle BCM is equal to the angle MAD as criss-crossing, then the triangles BMC and DMA are similar in two angles.
Since AM / MC = 3/2, the coefficient of similarity of triangles is 3/2.
The ratio of the areas of similar triangles is equal to the square of the similarity coefficient.
Sdma / Svms = (3/2) 2 = 9/4.
Sdma = Svms * 9/4 = 8 * 9/4 = 18 cm2.
Answer: The area of the triangle DMA is 18 cm2.
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