#### Question

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

#### Solution

For ΔABC,

AC^{2} = AB^{2} + BC^{2}

(5)^{2} = (3)^{2} + (4)^{2}

Therefore, ΔABC is a right-angled triangle, right-angled at point B.

Area of ΔABC`= 1/2xxABxxBC=1/2xx3xx4=6 cm^2`

For ΔADC,

Perimeter = 2*s* = AC + CD + DA = (5 + 4 + 5) cm = 14 cm

*s* = 14/2 = 7 cm

By Heron’s formula,

`"Area of triangle "=sqrt(s(s-a)(s-b)(s-c))`

`"Area of "triangle ADC=[sqrt(7(7-5)(7-5)(7-4))]cm^2`

`=(sqrt(7xx2xx2xx3))cm^2`

`=2sqrt21 cm^2`

= (2 x 4.583) cm^{2}

= 9.166 cm^{2}

Area of ABCD = Area of ΔABC + Area of ΔACD

= (6 + 9.166) cm^{2}

= 15.166 cm^{2}

= 15.2 cm^{2} (approximately)