Question

In trapezoid ABCD with legs AB and CD, point O is the intersection of the diagonals and

OM is the median to leg CD in △COD. If ACOM=2 ft^2, find AABO.

Answer 1

`4\ ft^2`

Explanation:

1. Consider triangle COD. The area of this triangle is

`A_(\triangle COD)=(1)/(2)\cdot CD\cdot h_O=(1)/(2)\cdot 2CM\cdot h_O=2\cdot (1)/(2)\cdot CM\cdot h_O=2A_(COM)=2\cdot 2=4\ ft^2.`

2. Consider triangles AOB and COD:

`A_(\triangle AOB)=(1)/(2)\cdot AO\cdot BO\cdot \sin \angle AOB,\\ \\A_(\triangle COD)=(1)/(2)\cdot CO\cdot DO\cdot \sin \angle COD`

and

`(AO)/(CO)=(DO)/(BO)\Rightarrow AO\cdot BO=CO\cdot DO` (triangles BOC and AOD are similar).

Since angles AOB and COD are vertical, then
`\sin \angle AOB=\sin \angle COD.`

Now,

`A_(\triangle AOB)=(1)/(2)\cdot AO\cdot BO\cdot \sin \angle AOB=(1)/(2)\cdot CO\cdot DO\cdot \sin \angle COD=A_(\triangle COD)=4\ ft^2.`