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A park, in the shape of a quadrilateral ABCD, has ∠ C = AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
Asked by Neha Last Modified
Answer 
Tasneem
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
For ΔABD,
By Heron’s formula,
Area of triangle
Area of ΔABD 
Area of the park = Area of ΔABD + Area of ΔBCD
= 35.496 + 30 m2 = 65.496 m2 = 65.5 m2 ( Approximate Value)
read lessLet us join BD.
Now BCD is a right triangle, right angled at C.
∴ area of triangle BCD = ½bh
= 1/2 × 12 × 5 = 6 × 5 = 30m².
By Pythagoras theorem, BD² = 12² + 5²
⇒ BD² = 144 + 25 = 169
∴ BD = √169 = 13m.
Now, area of triangle ABD
= √ 15 ( 15 - 13 ) ( 15 - 8 ) ( 15 - 9
= √ 15 × 6 × 2 × 7 = √ 5 × 3 × 3 × 2 × 2 × 7
= 6√35m² = 6 × 5.9 = 35.4m².
∴ the area of the park ABCD
= area of ABD + area of BCD
= 30m² + 35.4m² = 65.4m².
read less
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