# Find the area of a quadrilateral $\mathrm{ABCD}$ in which $\mathrm{AB}=3 \mathrm{~cm}, \mathrm{BC}=4 \mathrm{~cm}, \mathrm{CD}=4 \mathrm{~cm}$, $\mathrm{DA}=5 \mathrm{~cm}$ and $\mathrm{AC}=5 \mathrm{~cm}$.

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Given:

The quadrilateral $ABCD$ has $AB=3\ cm$, $BC=4\ cm$, $CD=4\ cm$, $DA=5\ cm$ and $AC=5\ cm$.

To do:

We have to find the area of the quadrilateral $ABCD$.

Solution:

In $\triangle ABC$, using Pythagoras theorem,

$AC^2=AB^2+BC^2$

$5^2=3^2+4^2$

​$25=25$

Therefore,

Area of $\triangle ABC=\frac{1}{2}\times3\times4=6\ cm^2$

In $\triangle ACD$,

$s=\frac{1}{2}(5+5+4)$

$=\frac{14}{2}$

$=7\ cm$

Area of $\triangle ACD=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{7(7-5)(7-5)(7-4)}$

$=\sqrt{(7)(2)(2)(3)}$

$=2\sqrt{21}\ cm^2$

$=9.17\ cm^2$

Therefore,

Area of quadrilateral $ABCD=$ Area of $\triangle ABC+$ Area of $\triangle ACD$

Therefore,

Area of quadrilateral $ABCD=6\ cm^2+9.17\ cm^2$

$=15.17\ cm^2$.

The area of a quadrilateral $ABCD$ is $15.17\ cm^2$.

Updated on 10-Oct-2022 13:42:04