433 looks good to me.

To hide:

square bracket hide="optional message in box" square bracket "what you want to hide here" then end with square bracket /hide square bracket.

For me, the key to the solution was when I drew the lines AE and CF. The software I use (Sketchpad) lets me put on lines that are truly parallel and maintains all the 90 angles. So I was able to experiment by varying the parallels. It looked like AE = BF so I thought I'd try to prove it. Then everything fell into place nicely.

With geometry I usually find the diagram is the key thing. Get that right and the route to an answer often becomes visible. I needed to put on two lines to show the 12 and 17 distances and it seemed it would keep things simpler if I drew these from A and from C.

Hope that helps.

Bob

]]>thanks a ton I got 433 (how to hide?) after you guys

I was clueless!

EDIT:

im a bob to im robert!]]>

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I did it with:

Bob

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Bob

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Welcome to the forum.

You'll get picture posting rights when you've been a member for a while.

Hopefully the one below is your problem.

Look at the angles I've marked x and y. They add to 90 and you'll see other angles of the same size in the diagram too.

So show that triangles ABE and BCF have the same angles and one side equal so they are congruent.

Then use Pythagoras to get a side.

Bob

]]>ABCD is a square. Parallel lines m, n, and p pass through vertices A, B, and C, respectively. The distance between m and n is 12, and the distance between n and p is 17. Find the area of square ABCD.

Im trying to post an image but it wont let im new

thanks I spent I couple hours on this and have no idea.

-thedarktiger]]>