PQ is the perpendicular bisector of side AD in trapezoid ABCD with AB perpendicular to BC and BC perpendicular to CD. If AB = 9, BC = 8, and CD =7, find area of APQB.
Can anyone help me out with this?
Hint: I would start with the perpendicular on AB through D. This establishes, if nothing else, two similar triangles on AD. Look carefully, they are not oriented the same so it might be hard to see. The little one has a leg on AD. AD is the hypotenuse on the other.
No, there's an easier solution to that.
Connect B and D and you will see two triangles, Triangle ABD and Triangle CDB.
Since PQ is parallel to AB and DC, we can see it creates two similar triangles for each triangle.
Let's say the intersection of PQ and BD is M.
Then, Triangle DMP is similar to Triangle DBA (1:2).
So, PM must be half the length of AB, which is 4.5
Do the same for the Triangle CDB, and you will see that MQ is half of CD, which is 3.5
So now we can find out that PQ equals PM + MQ = 4.5 + 3.5 = 8.
There you go.