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

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 = $$9.2{cm}^2 + 6{m}^2 = 15.2{cm}^2$$