The area of a triangle is one-third of the area of a parallelogram whose base is 24 cm.

Given:
Sides of triangle = 22 cm, 18 cm, 24 cm

Semi-perimeter
$s = \frac{22 + 18 + 24}{2} = 32$

Area of triangle (Heron’s formula)

$A = \sqrt{s(s-a)(s-b)(s-c)}$

$= \sqrt{32(32-22)(32-18)(32-24)}$

$= \sqrt{32 \times 10 \times 14 \times 8}$

$= \sqrt{256 \times 140}$

$= 16\sqrt{140}$

$= 16\sqrt{4 \times 35}$

$= 32\sqrt{35}$

Using $\sqrt{35} = 5.9$

Area of triangle
$= 32 \times 5.9 = 188.8 \text{ cm}^2$

Given:
Area of triangle = $\frac{1}{3}$ × Area of parallelogram

Area of parallelogram
$= 3 \times 188.8 = 566.4 \text{ cm}^2$

Area of parallelogram
$= \text{base} \times \text{height}$

$566.4 = 24 \times h$

$h = \frac{566.4}{24} = 23.6$

Height of the parallelogram = 23.6 cm