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ABCD is a plate in the shape of a parallelogram. EF is the line parallel to DA and passing through the point of intersection O of the diagonals AC and BD. Further, E lies on DC and F lies on AB. The triangular portion DOE is cut out from the plate ABCD. What is the ratio of area of remaining portion of the plate to the whole?

ABCD is a plate in the shape of a parallelogram. EF is the line parallel to DA and passing through the point of intersection O of the diagonals AC and BD. Further, E lies on DC and F lies on AB. The triangular portion DOE is cut out from the plate ABCD. What is the ratio of area of remaining portion of the plate to the whole? Correct Answer <span class="math-tex">\(\frac{7}{8}\)</span>

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If we suppose the complete area of the parallelogram is 8 units.

Then, Diagonal BD will divide it in 2 equal parts:

So, Area ABD = Area BCD = 4 units

Further CO will divide triangle BCD in 2 equal parts:

So, Area DOC = 4/2 = 2 units

Now,

Given EF || AD so, AD = EF

EO will divide DOC in 2 equal parts;

∴ Area of triangle DOE = 2/2 = 1 unit

When DOE is cut out from the plate ABCD;

∴ Ratio of area of remaining portion of the plate to the whole = (8 – 1) : 8 = 7 : 8

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