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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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ABCD is the park as shown in the figure. 

Join BD. 

In ∆DBC, we have 

DB2 = BC2 + CD2 [Pythagoras theorem] 

⇒ DB2 = (12)2 + 52 

⇒ DB = 144 + 25 = 169

⇒ DB = 13 m. 

Area of ∆DBC = 1/ 2 × base × height 

= 1/ 2 × 12 × 5 m2 = 30 m2

In ∆ABD, a = 9 m, b = 8 m, c = 13 m 

∴ s = (a +b + c) /2 =  (9+ 8+ 13) / 2 m = 15 m

∴ Area of the park = area of ∆DBC + area of ∆ABD 

= (30 + 35.5) m2 = 65.5 m2 Ans.

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