Question

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

Answer 1

Find the area of given quadrilateral

Given, a quadrilateral $ABCD$,

Here, for $∆ABC$

$AB=3cm$

$BC=4cm$

$AC=5cm$

The other side of the quadrilateral is

$CD=4cm$ and $DA=5cm$

To find the area of quadrilateral $ABCD$, we can add the areas of $∆ABC$ and $∆ACD$.

So,

$$\begin{gathered} \mathrm{Area}\mathrm{of}∆\mathrm{ABC}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height} \\ =\frac{1}{2}\times \mathrm{AB}\times \mathrm{BC} \\ =\frac{1}{2}\times 3\times 4 \\ =6c{m}^2 \end{gathered}$$

Now, for $∆ACD$

$AC=5cm$

$CD=4cm$

$DA=5cm$

Therefore, semi perimeter $S$ IS

$$\begin{gathered} S=\frac{4+5+5}{2} \\ =7cm \end{gathered}$$

Area of triangle $∆ACD$

$$\begin{gathered} =\sqrt{S(S-AC)(S-CD)(S-DA)} \\ =\sqrt{7(7-5)(7-4)(7-5)} \\ =2\sqrt{21} \\ =2\times 4.6\left(\mathrm{Approx}\right) \\ =9.2c{m}^2\left(\mathrm{Approx}\right) \end{gathered}$$

So, the area of quadrilateral will be,

$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{quadrilateral}=\mathit{∆}ABC\mathit{+}\mathit{∆}ACD \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}=6+9.2 \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}=15.2{\mathrm{cm}}^2 \end{gathered}$$

Hence, the area of quadrilateral is $15.2c{m}^2$