Answer :

In the quadrilateral ABCD,
we have, $\mathrm{?}ABC$,

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


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

= $3^2+4^2=9+16=25$
$?$ ${AC}^2=5^2$
$?$ AC = 5cm

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

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

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

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

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

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

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

Area of quadrilateral ABCD = Area of $\mathrm{?}ABC$ + Area of $\mathrm{?}ACD$

Hence from (ii) and (iii),

Area of quadrilateral ABCD = $9.2{cm}^2+6m^2=15.2{cm}^2$