Area of a trapezium using similar shapes

Notice that....

area of trapezium ABDE  =  area of triangle ACE  -  area of triangle BCD

And we know that △ACE and △BCD are similar triangles, because

m∠ACE  =  m∠BCD     and     m∠BDC  =  m∠AEC  =  90°

so by the AA similarity theorem, we can say that  △ACE ~ △BCD .

Now we can find the length of CE .

$\frac{\text{CE}}{\text{CD}}\,=\,\frac{\text{AE}}{\text{BD}}\\~\\ \frac{\text{CE}}{8\text{ cm}}\,=\,\frac{9\text{ cm}}{6\text{ cm}}\\~\\ \text{CE}\,=\,\frac{9\text{ cm}}{6\text{ cm}}\cdot8\text{ cm}\\~\\ \text{CE}\,=\,12\text{ cm}$

Now we can find the area of  △ACE  and the area of  △BCD .

area of △ACE  =  (1/2)( CE )( AE )

area of △ACE  =  (1/2)( 12 cm )( 9 cm )

area of △ACE  =  54 cm²

area of △BCD  =  (1/2)( CD )( BD )

area of △BCD  =  (1/2)( 8 cm )( 6 cm )

area of △BCD  =  24 cm²

Now we can find the area of trapezium ABDE.

area of trapezium ABDE  =  area of △ACE  -  area of △BCD

area of trapezium ABDE  =  54 cm²  -  24 cm²

area of trapezium ABDE  =  30 cm²