Find the area of the figure formed by joining the mid-points of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm.

Given: ABCD is a rhombus with diagonals 12 cm and 16 cm.

Construction: Join the midpoints of AB, BC, CD and DA of the rhombus ABCD and name them M, N, O and P, respectively, to form the quadrilateral MNOP.

Join the line PN.

To Prove: MNOP is a parallelogram.

Proof: The midpoint theorem states that the line segment joining the midpoints of two sides of a quadrilateral is parallel to the opposite side and half of its length.

Since M and N are midpoints of AB and BC, by midpoint theorem:

MN ∥ AC and MN = $\dfrac{1}{2}$ AC ……………..(1)

Similarly, P and O are midpoints of AD and DC. By the midpoint theorem:

PO ∥ AC and PO = $\dfrac{1}{2}$ AC …………………(2)

From equations (1) and (2), we get:

MN ∥ AC and MN = PO

Now, M and P are midpoints of AB and AD. By the midpoint theorem:

MP ∥ BD and MP = $\dfrac{1}{2}$ BD ……………..(3)

Similarly, N and O are midpoints of BC and DC. By the midpoint theorem:

NO ∥ BD and NO = $\dfrac{1}{2}$ BD …………………(4)

From equations (3) and (4), we get:

MP ∥ NO and MP = NO

Thus, quadrilateral MNOP is a parallelogram.

If a triangle and a parallelogram are on the same base and between the same parallels, the area of the triangle is equal to one-half area of the parallelogram.

Ar(Δ MNP) = $\dfrac{1}{2}$ Ar(∥gm ABNP) ……………..(5)

Ar(Δ PON) = $\dfrac{1}{2}$ Ar(∥gm PNCD) ……………..(6)

Then area of rhombus ABCD = $\dfrac{1}{2}$ x d₁ x d₂

= $\dfrac{1}{2}$ x (12 x 16) cm²

= 96 cm²

Since the parallelogram MNOP consists of two such halves:

Ar(MNOP) = $\dfrac{1}{2}$ (Ar(ABCD))

= $\dfrac{1}{2}$ x 96

= 48 cm²

Hence, the area of the figure formed = 48 cm².