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.

Answer 1

For ΔABC,

AC² = AB² + BC²

(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 = 2s = 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) cm²

                            = 9.166 cm²

Area of ABCD = Area of ΔABC + Area of ΔACD

                    = (6 + 9.166) cm²

                    = 15.166 cm²

                    = 15.2 cm² (approximately)