ABCD is a parallelogram with side AB = 10 cm. Its diagonals AC and BD are of length 12 cm and 16 cm respectively. Find the area of the parallelogram ABCD.

Let ABCD be a parallelogram with diagonals intersecting at O.

Since, diagonals of a parallelogram bisect each other.

∴ AO = $\dfrac{12}{2}$ = 6 cm and OB = $\dfrac{16}{2}$ = 8 cm.

In triangle AOB,

Let AB = a = 10 cm, BO = b = 8 cm and OA = c = 6 cm.

We know that,

Semi-perimeter (s) = $\dfrac{a + b + c}{2}$

= $\dfrac{10 + 8 + 6}{2} = \dfrac{24}{2}$ = 12 cm.

By Heron's formula,

$\text{Area of triangle } = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] = \sqrt{12(12 - 10)(12 - 8)(12 - 6)} \\[1em] = \sqrt{12 \times 2 \times 4 \times 6} \\[1em] = \sqrt{576} \\[1em] = 24 \text{ cm}^2.$

Since, diagonals of parallelogram intersect each other so O is the mid-point of BD.

∴ AO is the median of the △ABD.

Since, median divides the triangle into two triangles of equal area.

∴ Area of △AOB = $\dfrac{1}{2}$ × Area of △ABD ……(1)

Since, diagonal of a parallelogram divides it into two triangles of equal area.

∴ Area of △ABD = $\dfrac{1}{2}$ × Area of || gm ABCD.

Substituting above value of △ABD in equation 1 we get,

Area of △AOB = $\dfrac{1}{2} \times \dfrac{1}{2}$ Area of || gm ABCD

Substituting values in above equation we get,

24 = $\dfrac{1}{4}$ Area of || gm ABCD

⇒ Area of || gm ABCD = 24 × 4 = 96 cm².

Hence, area of || gm ABCD = 96 cm².