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

Let us given that, 

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

From above information we draw the figure, 

we connected B and D such the the triangle is formed

So, the triangle BDC is a right angled triangle

By using the Pythagoras theorem

we get,

BD² = BC² $+$ CD²

BD² = 12² $+$ 5²

BD² = 144 $+$ 25

BD = $\sqrt{169}$

BD = 13 m

Area of quadrilateral ABCD = area of ∆BCD$+$ area of ∆ABD

Now, Area of ∆BCD = 1/2 $\times$ base $\times$ height

= 1/2 $\times$ 12 m $\times$ 5 m

= 30 m²

Now, in ∆ABD, AB = a = 9 m, AD = b = 8 m, BD = c = 13 m

Semi Perimeter of ΔABD

s = (a $+$ b $+$ c)/2

= (9 $+$ 8 $+$ 13)/2

= 30/2

= 15 m

We know that, Heron's formula

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

where, s is the semi-perimeter = half of the perimeter

and  a, b and c are the sides of the triangle 

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

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

= $\sqrt{15 \times 6 \times 7 \times 2}$

= 6$\sqrt{35}$

= 35.5 m² 

Area of  triangle ABD = 35.5 m²

Therefore,

Area of park ABCD = 30 m² $+$ 35.5 m² = 65.5 m²

Hence,  the park ABCD occupies an area of 65.5 m².