Question

What is the figure formed by vertices ABCD in the figure given below and find its Area;

Answer 1

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if,

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In given figure,

ABCD is a quadrilateral and P,Q,R & S are mid-pints of sides AB,BC,CD and DA respectively. Then, PQRS is a square.

∴PQ=QR=RS=PS --------- (1)

and PR=SQ

But PR=BC and SQ=AB

∴AB=BC

Thus, all sides of quadrilateral ABCD are equal.

Hence, quadrilateral ABCD is either a square or a rhombus.

Now, in △ADB,

By using Mid-point theorem,

SP∣∣DB;SP= 1/2DB ------ (2)

Similarly in △ABC,

PQ∣∣AC;PQ= 1/2AC ----- (3)

From equation (1),

PS=PQ

From (2) and (3),

1/2 DB= 1/2 AC

∴DB=AC

Thus, diagonals of ABCD are equal and therefore quadrilateral ABCD is a square. So, diagonals of quadrilateral also perpendicular.

Answer 2

The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if,

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