Keeping a sense of proportion
We're renovating a garden shed which once doubled as a kind of shed/conservatory. Properly called a potting shed, perhaps? Anyhow, I'm trying to plan the proportions, before buying timber and perspex...
The right hand side is what's left of the shed bit, and is exactly 6' high. The base is 4'. I want the left hand side to be 4' high because that's the size the sheets of perspex come in (6'x4', actually). The question is, how long is that diagonal?
The squaw on the hippopotamus is twice the sum of the squaws on the other two hides.
ReplyDeleteI hope that helps.
4.4721 if I my memory above serves me correctly.
ReplyDeletehttp://mathforum.org/~sarah/hamilton/ham.hypotenuse.html
ReplyDeleteIn case you wanted to check. Don't blame me if the thing don't fit together!
Leave the squaws alone. Your better off without them. Never made much sense to me at school - I think I missed a bit out.
ReplyDeleteI needed to check up myself. Here is the aid memoir I was trying to remember! Hope you appreciate the irony in that -
ReplyDeleteTwo young braves and three squaws are sitting proudly side by side. The first squaw sits on a buffalo skin with her 50 pound son. The second squaw is on a deer skin with her 70 pound son. The third squaw, who weighs 120 pounds, is on a hippopotamus skin. Therefore, the squaw on the hippopotamus is equal to the sons of the squaws on the other two hides.
I think Pythagoras was easier.
ReplyDeleteBut that's where I went wrong in the pub on Sunday: the squaw on the hippopatumus was in the back of my mind, but here we have a quadrilateral, rather than a right angled triangle, so all bets are off.
Actually, I think the front bit and the the diagonal will both be as near 4' as makes no difference.
I'll blog a photo when it's done.