Question

10) If Diagonals AC and BD of a quadrilateral ABCD intersect at 0 in such a way that ar( AA0D ) artABOC). Then ABCD a. parallelogram b. square c. tripezium d. rectangle​

Answer 1

Answer:

Here, ABCD is a quadrilateral where diagonals AC and BD intersect at O.

⇒ ar(△AOD)=ar(△BOC)

Adding ar(ODC) on both sides,

⇒ ar(AOD)+ar(ODC)=ar(BOC)+ar(ODC)

⇒ ar(ADC)=ar(BDC)

Now, △ADC and △BDC lie on the same base DC and are equal in area and they lie between the lines AB and DC

⇒ AB∥DC ( Two triangles having the same base and equal areas lie between the same parallels )

In ABCD,

⇒ AB∥DC

So, one pair of opposite sides is parallel,

∴ ABCD is trapezium.

Step-by-step explanation:

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