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?

In the quadrilateral ABCD,
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 quadrilateral ABCD
= Area of \(\triangle{ABD}\) + Area of \(\triangle{BCD}\)

Hence from (i) and (ii),

Area of quadrilateral ABCD = \(35.4{m}^2 + 30{m}^2 = 65.4{m}^2\)