ABCD is a rhombus with AC = 12 cm and BD =16 cm.

P, Q, R and S are the mid-points of the sides AB, BC, CD and AD respectively.

We have to find the area of the quadrilateral PQRS formed by joining the mid-points of the rhombus ABCD

InΔ ABD, S is the mid-point of AD and P is the mid-point of AB.

⇒ SP || BD

And, SP = 1/2 of BD (∴ Mid-point theorem)

⇒ 1/2*16

= 8 cm

So, SP = 8 cm

InΔ BCD, R is the mid-point of CD and Q is the mid-point of BC.

⇒ QR || BD and QR = 1/2 of BD

⇒ 1/2*16

= 8 cm

So, QR is 8 cm

SP || BD and QR || BD⇒ SP || QR

Also, SP = QR = 8 cm

⇒ SPQR is a parallelogram.

Similarly, PQ = SR = 1/2 of AC

⇒ 1/2*12

= 6 cm

So, PQ = SR = 6 cm

Now,

BD || SP⇒ SN || OM

AC || RS⇒ ON || MS

⇒ NOMS is a parallelogram.

∠ AOD = 90° (∴ Diagonals of a parallelogram bisect at right angle)

⇒∠ MSN = 90° (∴ NOMS is a parallelogram)

⇒ PQRS is a rectangle.

Area of rectangle PQRS = SP× PQ

⇒ 8× 6

= 48 sq cm

So, area of the quadrilateral formed by joining the mid-points of the sides of

rhombus ABCD is 48 sq cm.