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# A park, in the shape of a quadrilateral ABCD, has $$\angle{C}$$ = $$90^\circ$$ , AB = 9m, BC = 12m, CD = 5m and AD = 8m. How much area does it occupy?

we have, right angled triangle $$\triangle{BCD}$$,

We have,
$${BD}^2 = {BC}^2 + {CD}^2$$
(By Pythagoras theorem)

= $${12}^2 + {5}^2 = 144 + 25 = 169$$
$$\Rightarrow$$ $${BD}^2 = {13}^2$$
$$\Rightarrow$$ BD = 13m

Also, In $$\triangle{ABD}$$, we have,

AB = 9m, BD = 13m, DA = 8m

Now, we know that,
$$s = \frac{a + b + c}{2}$$
$$\therefore$$ $$s = \frac{9 + 13 + 8}{2} = \frac{30}{2} = 15m$$

Now, Area of triangle
= $$\sqrt{15(15 - 9)(15 - 13)(15 - 8)}$$
(Since, Heron's formula [area = $$\sqrt{s(s - a)(s - b)(s - c)}$$])

= $$\sqrt{15 × 6 × 2 × 7}$$
= $$\sqrt{3 × 5 × 3 × 2 × 2 × 7}$$
= $$3 × 2\sqrt{5 × 7} {m}^2$$
= $$6\sqrt{35} {m}^2$$ = 6 × 5.9$${m}^2$$ = 35.4$${m}^2$$(approx) ...(i)

Since, $$\triangle{BCD}$$ is an right angled triangle,

Area of $$\triangle{BCD}$$ = $$\frac{1}{2}$$ × BC × CD
(Since, Area of triangle = $$\frac{1}{2}$$ × Base × Height)

= $$\frac{1}{2} × 12 × 5 = 30{m}^2$$ ...(ii)

= Area of $$\triangle{ABD}$$ + Area of $$\triangle{BCD}$$
Area of quadrilateral ABCD = $$35.4{m}^2 + 30{m}^2 = 65.4{m}^2$$