Question

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 ?​

Answer 1

Answer:

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Answer 2

Answer:

Here, ABCD is a parallelogram in which its diagonals AC and BD intersect each other in O.  

∴  O is the mid - point of AC as well as BD.

Now, in △ADB , AO is its median  

∴ ar(△ADB) = 2 ar(△AOD) [∵ median divides a triangle into two triangles of equal areas]

So, (△ADB) = 2 × 4 = 8 cm2    

Now, △ADB and ||gm ABCD lie on the same base AB and lie between same parallel AB and CD .

∴ ar(ABCD) = 2 ar(△ADB)  

= 2 × 8  

= 16 cm2