Determine lengths and angles in special quadrilaterals Geometry

The quadrilateral is a trapezoid. So drawing the altitude \overline{DH}, the quadrilateral HBCD is a rectangle:

Hence:

DH =BC =3

And:

BH = CD =2

Also:

AH = 6-BH = 4

Using the Pythagoras theorem, we have:

a= \sqrt{3^2+4^2}=5

The quadrilateral is a rhombus. Thus:

DC =AC =5

And:

OC= BO=3

Using the Pythagoras theorem, we have:

a= \sqrt{5^2+3^2}=4

The quadrilateral AHCH' is a rectangle with heigth \overline{DH'} :

Hence:

AH' =BC =5

We know that:

\sin{\widehat{a}} =\dfrac{AH'}{AD} = \dfrac{5}{10}= \dfrac{1}{2}

Therefore:

\widehat{a}=30^\circ

The quadrilateral is a square. Thus:

BD = CD =a

Using the Pythagoras theorem, we have:

\left(BC\right)^2={a^2+a^2}

8^2 = 2a^2

a^2= \dfrac{64}{2}=32

a=\sqrt{32}=4\sqrt{2}

The quadrilateral is a kite. Thus:

CD=BC =5

The triangle BCD is a equilateral. Therefore:

\widehat{CDB} = 60^\circ

And:

\widehat{ADB} = 130^\circ -\widehat{CDB} = 130^\circ-60^\circ = 70^\circ

Since ABCD is a quadrilateral, we have:

\widehat{A} + \widehat{B} + \widehat{C}+ \widehat{D} = 360^\circ

\widehat{A} = 360^\circ- \widehat{B} - \widehat{C}- \widehat{D}

\widehat{A} = 360^\circ- 130^\circ - 60^\circ- 130^\circ

\widehat{A} = 40^\circ

The quadrilateral is a rectangle. Thus:

CD=AB =a

We know that:

\cos\left(\widehat{DAC}\right) = \dfrac{AD}{AC}

\cos\left(60^\circ\right) = \dfrac{4}{AC}

\dfrac{1}{2}= \dfrac{4}{AC}

AC=8

Using the Pythagoras theorem, we have:

a= \sqrt{8^2-4^2}=\sqrt{48}=4\sqrt{3}

The quadrilateral is a rhombus so we have:

BC =DC =5

The triangle BDC is isosceles. Therefore:

\widehat{a} = \widehat{BDC}

Hence:

\widehat{a}+\widehat{a}+60^\circ=180^\circ

2\widehat{a}=120^\circ

\widehat{a}=60^\circ