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Updated on : 2022-09-05

Solution

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Correct option is D)

$PQRS$ is a quadrilateral.

$[PA=AQ,QB=BR,RC=CS,SD=DP.]$ ....(1)

Let $ABCD$ be a rhombus.

To find out:

The kind of the quadrilateral $PQRS$

Construction:

We join $AC&BD$, they intersect at $O$.

Solution:

$∵ABCD$ is a rhombus and its diagonals intersect at $O$

$∴[DO=BO,AO=CO,∠DOC=∠BOC=∠DOA=∠BOA=90_{o}]$ ....(2)

Now, between $DOCS$ and $OCBR$, we have

$DO=BO$ (by 2) ,

$OC$ common side, $RC=CS$ (by 1 ), $∠DOC=∠BOC=90_{o}$

$∴DOCS$ and $OCBR$ are identical rectangles.

Similarly, it can be proved that all the four sub figures of $PQRS$ are identical rectangles.

$∴PQRS$ is a rectangle.

So its diagonals are equal.

Hence, option D.

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