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

Answer:

$4\ ft^2$

Step-by-step explanation:

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

$A_{\triangle COD}=\dfrac{1}{2}\cdot CD\cdot h_O=\dfrac{1}{2}\cdot 2CM\cdot h_O=2\cdot \dfrac{1}{2}\cdot CM\cdot h_O=2A_{COM}=2\cdot 2=4\ ft^2.$

2. Consider triangles AOB and COD:

$A_{\triangle AOB}=\dfrac{1}{2}\cdot AO\cdot BO\cdot \sin \angle AOB,\\ \\A_{\triangle COD}=\dfrac{1}{2}\cdot CO\cdot DO\cdot \sin \angle COD$

and

$\dfrac{AO}{CO}=\dfrac{DO}{BO}\Rightarrow AO\cdot BO=CO\cdot DO$ (triangles BOC and AOD are similar).

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

Now,

$A_{\triangle AOB}=\dfrac{1}{2}\cdot AO\cdot BO\cdot \sin \angle AOB=\dfrac{1}{2}\cdot CO\cdot DO\cdot \sin \angle COD=A_{\triangle COD}=4\ ft^2.$