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This is a quantative comparison question. It has a diagram with it that I will have to describe (its pretty easy).

Draw a right triangle with unequal legs. Now, draw a square with the hypotenues of the triangle as one of its sides. Thats the diagram.

Question: Which is greater, the area of the square or four times the area of the triangle?

By Serene (Serene) on Friday, August 15, 2003 - 04:34 pm: Edit

area of square. i'll make a drawing.

By Jason817 (Jason817) on Friday, August 15, 2003 - 04:36 pm: Edit

It's A I think.

If it were a 3-4-5 right triangle, then the area of the square (25) would be greater than 4x the area (4 x 6 = 24).

If it were a 1-rt 3-2 triangle, the area of the square is 4 and the area of the triangle is 3.46.

By Serene (Serene) on Friday, August 15, 2003 - 04:38 pm: Edit

Testtaker: see the picture in my profile. The square is greater than the four triangles by the smaller square in the middle.

By Jason817 (Jason817) on Friday, August 15, 2003 - 04:40 pm: Edit

Wow. Never thought of it that way.

By Testtaker (Testtaker) on Friday, August 15, 2003 - 04:46 pm: Edit

serene, the side of the square is the hypotenues of the triangele, so your diagram is not what I meant

if you pretend that thing on top of the triangle is a square and mentally remove all the dots, this is what it looks like:

.../*
../...*
./.......*
|*....../
|..*.../
|___*

By A87 (A87) on Friday, August 15, 2003 - 04:47 pm: Edit

Wow, that's very smart Serene. I had to resort to plugging in numbers, which is risky in quantitive comparison:

4(1/2bh) = 2bh

hyp^2 = area of square
h^2 + b^2 = hyp^2
h^2 + b^2 = area of square

A. h^2 + b^2
B. 2bh

By Xiggi (Xiggi) on Friday, August 15, 2003 - 04:49 pm: Edit

It is also a good graphic illustration of Pyth Theorem.

By Fairyofwind (Fairyofwind) on Friday, August 15, 2003 - 04:52 pm: Edit

With a given hypotenuse, the vertices opposite the hypotenuse for all right triangles with that hypotenuse trace a circle. Consider the two points on the circle such that when a line is drawn perpendicular to the hypotenuse it passes through the midpoint of the hypotenuse. In this case, the height of the triangle with respect to the hypotenuse is greatest (= radius of circle = 1/2 hypotenuse length), yet since the triangle has unequal legs, the area of the triangle must be less than (1/2)(h/Sqrt[2])^2=h^2/4, whereas the area of the square is always h^2. So 4 times the area of the triangle < h^2.

By Testtaker (Testtaker) on Friday, August 15, 2003 - 04:54 pm: Edit

Guys, I think you are all misinterpreting my question (serene's diagram is not the one in my book). Also, this answer the wrong answer to my question. Take a look at the diagram I attempted to make above (I know its bad-i'll try to make a better one).

By Serene (Serene) on Friday, August 15, 2003 - 04:55 pm: Edit

Testtaker: the bigger square. the left-most triangle is the given triangle, from its hypotenus you can see the big square. then inside the square i fitted in 4 congruent triangles with space (the smaller square) left.

By Fairyofwind (Fairyofwind) on Friday, August 15, 2003 - 04:56 pm: Edit

testtaker. Serene's diagram's smaller square degenerates into a point when the triangle is isosceles, but since the triangle has unequal legs, it never will, and thus the bigger square will always have a larger area than 4 times the area of the triangle.

By Serene (Serene) on Friday, August 15, 2003 - 04:58 pm: Edit

fairy: you know what software I really want? Geometer's Sketchpad. don't have the money to buy it though =P

By Xiggi (Xiggi) on Friday, August 15, 2003 - 05:05 pm: Edit

You do not need to make this so complicated. This is a QC on a SAT1. No need to come up with proofs, etc.

Like somoeone said. Plug the value for any right triangle. 3-4-5 is beautiful. Realize that 25>3.4/2*4 and .... move on. This is probably a level 1 question.

By Zerg_Vvins (Zerg_Vvins) on Friday, August 15, 2003 - 05:06 pm: Edit

Serene: try using mirc to get that software

By Testtaker (Testtaker) on Friday, August 15, 2003 - 05:06 pm: Edit

actually, it is the last question in the set!

By Xiggi (Xiggi) on Friday, August 15, 2003 - 05:11 pm: Edit

That looks like a cool program for 40 bucks.

By Xiggi (Xiggi) on Friday, August 15, 2003 - 05:14 pm: Edit

actually, it is the last question in the set!

Wow - considering how much they test the right triangles theories, this problem did not seem that hard for anyone who prepped just a bit.

By Fairyofwind (Fairyofwind) on Friday, August 15, 2003 - 05:20 pm: Edit

I have gsp 4... not by the most legal means...

By Tootall (Tootall) on Friday, August 15, 2003 - 05:23 pm: Edit

a is smaller side of right triangle. B is larger side of right triangle. a^2 +b^2=c^2 square root of a^2 +b^2=c c^2 is equal to the area of the square. squareroot of (a^2 +b^2)* squareroot of (a^2 +b^2) equals a^2 +b^2, this is equal to square. 4* area of triangles is equal to 2ab.

so (a^2 +b^2) and 2AB is what you're comparing

(a^2 +b^2) is greater since a and b can't be 1 or zero. It took a while, but it took me less than 30 seconds to solve. Serene made a nice diagram which should help you out.

By Fairyofwind (Fairyofwind) on Friday, August 15, 2003 - 05:43 pm: Edit

Good. or (a-b)^2>0, a^2-2ab+b^2>0, a^2+b^2>2ab.

By Raindrops (Raindrops) on Friday, August 15, 2003 - 06:32 pm: Edit

Testtaker, are you saying that the square is in the triangle? That's how I see it.