A park in the shape of a quadrilateral ABCD, has angle

Solution:

Given sides of a quadrilateral are AB = 9 m, BC = 12 m, CD = 5 m, DA = 8 m.

Let us join BD

In triangle BCD, apply Pythagoras theorem

$B D^{2}=B C^{2}+C D^{2}$

$B D^{2}=12^{2}+5^{2}$

BD = 13 m

Area of triangle BCD = 1/2 × BC × CD

= 1/2 × 12 × 5

$=30 \mathrm{~m}^{2}$

Now, in triangle ABD

Perimeter = 2s = 9 m + 8m + 13m

s = 15 m

By using Heron's Formula,

Area of the triangle $\mathrm{ABD}=\sqrt{\mathrm{s} \times(\mathrm{s}-\mathrm{a}) \times(\mathrm{s}-\mathrm{b}) \times(\mathrm{s}-\mathrm{c})}$

$=\sqrt{15 \times(15-9) \times(15-8) \times(15-13)}$

$=35.49 \mathrm{~m}^{2}$

Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD

$=(35.496+30) m^{2}$

$=65.5 \mathrm{~m}^{2}$