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Find the area of a quadrilateral $ABCD$ in which $AB = 3\ cm, BC = 4\ cm, CD = 4\ cm, DA = 5\ cm$ and $AC = 5\ cm$.
Given:
A quadrilateral $ABCD$ in which $AB = 3\ cm, BC = 4\ cm, CD = 4\ cm, DA = 5\ cm$ and $AC = 5\ cm$.
To do:
We have to find the area of the quadrilateral.
Solution:
$AC$ is the diagonal which divides the quadrilateral into two triangles.
In $\triangle ABC$,
$AB = 3\ cm, BC = 4\ cm, AC = 5\ cm$
Therefore,
$s=\frac{a+b+c}{2}$
$=\frac{3+4+5}{2}$
$=\frac{12}{2}$
$=6$
Area of the triangle $\mathrm{ABC}=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{6(6-3)(6-4)(6-5)}$
$=\sqrt{6 \times 3 \times 2 \times 1}$
$=\sqrt{36}$
$=6 \mathrm{~cm}^{2}$
In $\Delta \mathrm{ADC}$,
$\mathrm{AC}=5 \mathrm{~cm}, \mathrm{AD}=5 \mathrm{~cm}, \mathrm{CD}=4 \mathrm{~cm}$
$s=\frac{a+b+c}{2}$
$=\frac{5+5+4}{2}$
$=\frac{14}{2}$
$=7$
Area of $\Delta \mathrm{ADC}=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{7(7-5)(7-5)(7-4)}$
$=\sqrt{7 \times 2 \times 2 \times 3}$
$=2 \sqrt{21}$
$=2 \times 4.58$
$=9.16 \mathrm{~cm}^{2}$
Therefore,
Area of quadrilateral $\mathrm{ABCD}=6+9.16 \mathrm{~cm}^{2}$
$=15.16 \mathrm{~cm}^{2}$.
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