Question

A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

Answer 1

Let us join BD.

In ΔBCD, applying Pythagoras theorem,

BD² = BC² + CD²

= (12)² + (5)²

= 144 + 25

BD² = 169

BD = 13 m

Area of ΔBCD

`= 1/2xxBCxxCD = (1/2xx12xx5)m^2=30m^2`

For ΔABD,

`s="Perimeter"/2=(9+8+12)/2=15m`

By Heron's formula,

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

`"Area of "triangleABD=[sqrt(15(15-9)(15-8)(15-13))]m^2`

                           `=(sqrt(15xx6xx7xx2))m^2`

                           `=6sqrt35 m^2`

                            = (6 x 5.916) m²

                            = 35.496 m²

Area of the park = Area of ΔABD + Area of ΔBCD

                        = 35.496 + 30 m² 

                        = 65.496 m²

                        = 65.5 m² (approximately)