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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?
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Solution
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)
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