The answers in the box are wrong and i looked through my book to see how to do it,i can't figure it out
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
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