Question

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

Answer 1

From above figure,

Area of quad $ABCD=ar\left(△ABC\right)+ar\left(△ACD\right)$

In $△ABC$,

$A{B}^2+B{C}^2=3^2+4^2=9+16=25=5^2=A{C}^2$

We know that, if sum of square of two sides is equal to the square of third side, then the triangle is right angled triangle.

So, By Pythagoras theorem, $△ABC$ is a right angled triangle.

$\Rightarrow$ Area of right-angled triangle $=\frac{1}{2}\times$ Base $\times$ Altitude

Now, $ar\left(△ABC\right)=\frac{1}{2}\times AB\times BC=\frac{1}{2}\times 3\times 4c{m}^2=6c{m}^2$

In $\Delta ACD$,

Here $a=B=5cm$ and $b=4cm$

$s=\frac{a+b+c}{2}$

$\Rightarrow s=\frac{5+5+4}{2}cm$

$\Rightarrow s=\frac{14}{2}cm$

$\Rightarrow s=7cm$

$\Rightarrow s-a=s-b=(7-5)cm$

$\Rightarrow s-a=s-b=2cm$

$\Rightarrow s-c=(7-4)cm$

$\Rightarrow s-c=3cm$

By Heron's formula,

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

$=\sqrt{7cm\times \left(2cm\right)\times \left(2cm\right)\times \left(3cm\right)}$

$=\sqrt{4\times 21}c{m}^2$

$=2\sqrt{21}c{m}^2$

Therefore, Area of $\Delta ACD=2\sqrt{21}c{m}^2$

$\therefore ar\left(ABCD\right)=ar\left(△ABC\right)+ar\left(△ACD\right)$

$=6c{m}^2+2\sqrt{21}c{m}^2$

$=(6+2\sqrt{21})c{m}^2$

Hence, the area of quadrilateral $ABCD$ is $(6+2\sqrt{21})c{m}^2$.