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#1 2007-01-26 14:08:22

efmchico
Member
Registered: 2007-01-15
Posts: 7

Geometry Construction I can't figure out

It says to construct an isosceles right triangle whose area is equal to the sum of the areas of the two given isosceles right triangles.

And there are two given isosceles right triangles, one smaller, one larger.

The teacher hinted to do some calculation but of course, the straight edge is unmarked, so i'm not sure how to even start the hint.

Any help is appreciated

Thanks

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#2 2007-01-27 02:14:26

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Geometry Construction I can't figure out

I think I've got a way to do this. It might not be the most efficient way, but I'm pretty sure it works.

I tried describing it but then it got really complicated and confusing, so I drew a picture instead. Unfortunately, it's in Paint, and for some reason the text thing wasn't working so I had to write in annotations myself. But hopefully you can see what's going on.

4efmchicoxt8.png

You basically measure the side of the smaller triangle (by setting the compass length), then extend the bigger triangle by that amount, then draw a line as shown in the diagram. From there, you draw a perpendicular of the same length as the line you just drew, and then you draw another line as shown.

The black lines show what you're given and the red lines show what you draw. This works because it's essentially the Pythagorean theorem divided by 2. Instead of squares, we're using exact halves of squares, but the principle is exactly the same. The triangle that you draw has a length of √(a²+b²), where a and b are the lengths of the other two sides.


Why did the vector cross the road?
It wanted to be normal.

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#3 2007-01-29 18:43:49

efmchico
Member
Registered: 2007-01-15
Posts: 7

Re: Geometry Construction I can't figure out

The triangle that you draw has a length of ?(a²+b²), where a and b are the lengths of the other two sides.

How do you get this?  Thanks.

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#4 2007-01-29 22:38:00

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Geometry Construction I can't figure out

By Pythagoras's Theorem. The first line you draw gets you two sides of a right-angled triangle with each side being the length of one of the given triangles, so then joining up the corners will get you the correct length.

The area of the two triangles together is a²/2 + b²/2, and the area of the triangle you draw is (√(a²+b²))²/2 = a²/2 + b²/2, so the areas are equivalent.

I think that this topic would actually be better off in Help, so I hope you still find it. smile


Why did the vector cross the road?
It wanted to be normal.

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