Find the area of a quadrilateral ABCD in which AB = 3cm, BC = 4cm, CD = 4cm, DA = 5cm and AC = 5cm.

In the quadrilateral ABCD,
we have, \(\triangle{ABC}\),

We have, AB = 3cm, BC = 4cm, CA = 5cm


We have, \({AC}^2 = {AB}^2 + {BC}^2\)
(By Pythagoras theorem)

= \({3}^2 + {4}^2 = 9 + 16 = 25\)
\(\Rightarrow \) \({AC}^2 = {5}^2\)
\(\Rightarrow \) AC = 5cm

Hence, \(\triangle{ABC}\) is a right triangle ...(i)

Now in \(\triangle{ABD}\), we have,
AC = 5cm, CD = 4cm, DA = 5cm
We know that,
\(s = \frac{a + b + c}{2}\)
\(\therefore \) \(s = \frac{5 + 4 + 5}{2} = \frac{14}{2} = 7cm\)

Now, Area of triangle
= \(\sqrt{7(7 - 5)(7 - 4)(7 - 5)}\)
(Since, Heron's formula [area = \(\sqrt{s(s - a)(s - b)(s - c)}\)])

= \(\sqrt{7 × 2 × 3 × 2}\)
= \(2\sqrt{7 × 3} {cm}^2\)
= \(2\sqrt{21} {cm}^2\) = 2 × 4.6\({cm}^2\) = 9.2\({cm}^2\)(approx) ...(ii)

Since, \(\triangle{ABC}\) is an right angled triangle, ...(from(i))

Area of \(\triangle{ABC}\) = \(\frac{1}{2}\) × AB × BC
(Since, Area of triangle = \(\frac{1}{2}\) × Base × Height)

= \(\frac{1}{2} × 3 × 4 = 6{cm}^2\) ...(iii)

Area of quadrilateral ABCD = Area of \(\triangle{ABC}\) + Area of \(\triangle{ACD}\)

Hence from (ii) and (iii),

Area of quadrilateral ABCD = \(9.2{cm}^2 + 6{m}^2 = 15.2{cm}^2\)