Find the area of quadrilateral ABCD in which AB = 29 cm, BC = 21 cm, AC = 20 cm, CD = 34 cm and DA = 42 cm.

From figure,

The diagonal AC divides the quadrilateral ABCD into two triangles : △ABC & △ACD

Area of △ABC :

Let sides AB = a = 29 cm, BC = b = 21 cm, AC = c = 20 cm

s = $\dfrac{29 + 21 + 20}{2}$

= $\dfrac{70}{2}$ = 35.

By formula,

$\Rightarrow \text{Area of triangle} = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] = \sqrt{35 \times (35 - 29) \times (35 - 21) \times (35 - 20)} \\[1em] = \sqrt{35 \times (6) \times (14) \times (15)} \\[1em] = \sqrt{44100} \\[1em] = 210 \text{ cm}^2.$

Area of △ACD

Let sides be AD = a = 42 cm, CD = b = 34 cm, AC = c = 20 cm

s = $\dfrac{42 + 34 + 20}{2}$

= $\dfrac{96}{2}$ = 48.

By formula,

$\Rightarrow \text{Area of triangle} = \sqrt{s(s - a)(s - b)(s - c)} \\[1em] = \sqrt{48 \times (48 - 42) \times (48 - 34) \times (48 - 20)} \\[1em] = \sqrt{48 \times (6) \times (14) \times (28)} \\[1em] = \sqrt{112896} \\[1em] = 336 \text{ cm}^2.$

Area of quadrilateral ABCD = Area of △ABC + Area of △ACD

= 210 + 336

= 546 cm².

Hence, area of quadrilateral = 546 cm².