Answer :

Given: A rectangle with sides 8cm and 6cm.

Consider the Rectangle ABCD

Here DR = RD = AP = PB = 8/2 = 4cm (∵ P and R are the midpoints of DC and AB respectively)

and AS = SD = BQ = QC = 6/2 = 3cm(∵ S and Q are the midpoints of AD and BC respectively)

Now, consider the ΔRSD

By Pythagoras theorem

SR^{2} = SD^{2} + DR^{2}

SR^{2} = 3^{2} + 4^{2}

SR^{2} = 9 + 16

SR^{2} = 25

SR = 5 cm

Similarly using Pythagoras theorem in ΔQRC , ΔPBQ and ΔAPS

We get RQ = QP = PS = 5cm

∴ SR = RQ = QP = PS = 5cm

∴ PQSR is Rhombus of side length 5cm

Area of the rhombus is given by:

∴ Area of the rhombus = = = 24cm^{2}

∴ Area(PQRS) = 24cm^{2}

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