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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Updated on: 10-Oct-2022

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