SOLUTION: In class we had to figure out how to solve problems bu using solving the sub-problem of that problem. Which is basically just steps to get to figuring out the big picture. In the f

Algebra ->  Equations -> SOLUTION: In class we had to figure out how to solve problems bu using solving the sub-problem of that problem. Which is basically just steps to get to figuring out the big picture. In the f      Log On


   

Question 231878: In class we had to figure out how to solve problems bu using solving the sub-problem of that problem. Which is basically just steps to get to figuring out the big picture. In the following problem I wasn't quite sure where to start with that. Here is the problem:
"In right triangle RED, angle R is the right angle. Point O is on segment ER and point A is on segment ED. Segment OA is perpendicular to segment ED. EA = 6, AD = 14, and ER = 16. Find the area of quadrilateral ROAD."
So with the picture you draw with that, do I have to cut the right triangle in half and then solve? I was so confused nor do I know where to start with finding the sub problems. If anyone has a suggestion it would be very helpful. Much appreciated in advance!
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!




Notice a few things right off:

1. EA = 6 and AD = 14 means that ED = 20.

2. OA perpendicular to RE means OA parallel to RD, hence:

* Triangle OEA similar to triangle RED

* Quadrilateral ROAD is a trapeziod

3. 20 is 4 times 5 and 16 is 4 times 4, hence RD has to be 12: 4 times 3 -- 3-4-5 right triangle.
(verify by using Pythagoras and calculating if you want)


The hypotenuse of OEA is 6, so because of similar triangles. Hence OE = 4.8 and OA = 3.6. Verification left as an exercise for the student.

Since OE = 4.8, RO = 16 - 4.8 = 11.2. And because of right angle R, RO is an altitude of the trapezoid

The formula for the area of a trapezoid is:



Plug in the values:



You can do your own arithmetic.

John