Question

EXERCISE 12.2
A park, in the shape of a quadrilateral ABCD, has 2 C = 90°, AB = 9 m, BC = 12 m,
CD=5 m and AD=8 m. How much area does it occupy?​

Answer 1

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Given: ∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m

Construction: Joim BD

In ΔBCD,

By applying Pythagoras theorem,

BD2 = BC2 + CD2

⇒ BD2 = 122 + 52

⇒ BD2 = 169whats

⇒ BD = 13 m

Area of ΔBCD = 1/2 × 12 × 5 = 30 m2

Now,

Semi perimeter of ΔABD(s) = (8 + 9 + 13)/2 m = 30/2 m = 15 m

Using heron's formula,

Area of ΔABD = √s (s-a) (s-b) (s-c)

= √15(15 - 13) (15 - 9) (15 - 8) m2

= √15 × 2 × 6 × 7 m2

= 6√35 m2 = 35.5 m2 (approx)

Area of quadrilateral ABCD = Area of ΔBCD + Area of ΔABD = 30 m2 + 35.5m2 = 65.5m2

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