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The diagonal lines AC and BD of isosceles trapezoid ABCD intersect at the two bottoms of O, OE parallel trapezoid, and the area of △ BOC is 16 square centimeter It's urgent. I'm begging you. Please
8 square centimeters
As shown in the figure, in the trapezoidal ABCD, ad ‖ BC, ad = 6cm, BC = 9cm, it is known that the area of △ ADC is 12cm2, take a point E on the extension line of Ba, and de ‖ AC, calculate the area of △ ABC and △ AED
ADCs △ ADCs △ ABC = ADBC (2 points) and ad = 6, BC = 9, and the area of △ ADC is 12 {12S △ ABC = 69 6569; s △ ABC = 18 (2 points) \\\\\\\\\\\\\\\\\\\\\adcs \\\\\\\\\\\\\\\\\\\\\\(2)
The area of rectangle ABCD is 10 square centimeters, the area of parallelogram bdef is () square centimeters, and the sum of triangle CDF and triangle CBF is () centigrade
Sorry, wrong input, the final should be square centimeter!
The area of the parallelogram is 10. The sum of the areas of the two triangles is 5. I worked it out. Children's homework is really difficult
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