Find the area of parallelogram, if its two adjacent sides are 12 cm and 14 cm, and if the diagonal connecting their ends is 18 cm.

By formula,

Area of triangle = $\sqrt{s(s - a)(s - b)(s - c)}$

In triangle ABC,

Let, AB = a = 12 cm, BC = b = 14 cm, AC = c = 18 cm

s = $\dfrac{a + b + c}{2} = \dfrac{12 + 14 + 18}{2}$ = 22 cm.

Substituting values we get :

$\text{Area of triangle ABC} = \sqrt{22 \times (22 - 12) \times (22 - 14) \times (22 - 18)} \\[1em] = \sqrt{22 \times 10 \times 8 \times 4} \\[1em] = \sqrt{7040} \\[1em] = 83.9 \text{ cm}^2.$

A diagonal of //gm divides it into triangles of equal area.

Thus, Area of //gm ABCD = 2.Area of triangle ABC

= 2 × 83.9

= 167.80 cm².

Hence, area of parallelogram = 167.80 cm².