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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?

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#### Solution

Let us join BD.

In ΔBCD, applying Pythagoras theorem,

BD^{2} = BC^{2} + CD^{2}

= (12)^{2} + (5)^{2}

= 144 + 25

BD^{2} = 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) m^{2}

= 35.496 m^{2}

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

= 35.496 + 30 m^{2 }

= 65.496 m^{2}

= 65.5 m^{2} (approximately)

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