Find the area of the quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
For triangle ABC
$A C^{2}=B C^{2}+A B^{2}$
25 = 9 + 16
So, triangle ABC is a right angle triangle right angled at point R
Area of triangle ABC = 12 × AB × BC
= 1/2 × 3 × 4
$=6 \mathrm{~cm}^{2}$
From triangle CAD
Perimeter = 2s = AC + CD + DA
2s = 5 cm+ 4 cm+ 5 cm
2s = 14 cm
s = 7 cm
By using Heron's Formula
Area of the triangle $\mathrm{CAD}=\sqrt{\mathrm{s} \times(\mathrm{s}-\mathrm{a}) \times(\mathrm{s}-\mathrm{b}) \times(\mathrm{s}-\mathrm{c})}$
$=\sqrt{7 \times(7-5) \times(7-4) \times(7-5)}$
$=9.16 \mathrm{~cm}^{2}$
Area of ABCD = Area of ABC + Area of CAD
$=(6+9.16) \mathrm{cm}^{2}$
$=15.16 \mathrm{~cm}^{2}$
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