<p>k, this is an actual question that is really tough. I've spent the last half hour trying to understand it while getting nowhere. </p>
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<p>I keep getting the answer as (m*squareroot of 3)/2 which is what I believe the triangle with base m/2 and hypotenuse m is, considering it is a 30/60/90. The answer however is A which would mean it has to be a 45-45-90 which makes no sense to me.</p>
<p>Would be great if someone could shed some light on this.</p>
<p>Yeah, that is a tough question! But not if we make up our own #s. H=one of the legs. Let's say that M=E=4. So now the point where the height hits the square to the base of the triangle is 2, or 1/2 of M(ONE OF THE OTHER LEGS). Now we're trying to find the hypotenuse. With all the sides being equal, it's an equilateral triangle, so all the angles are 60. So then bisect it to get a 30-60-90 triangle to get 2<em>SQ(3) as the hypotenuse. So we have 2 as the leg and 2</em>SQ(3) as the hypotenuse and H as a leg. Work it out to get SQ(8) as H. If you put that into your calculator, it comes out to be 2.8284. So now we said that M=4, so plug that in and A was the first one I got to be 2.8284(4/SQ(2).</p>
<p>Find an equation relating e, h, and m</p>
<p>The pythagoreon theorem is appropriate.</p>
<p>becuase the base is a square, you know the diagonal is M (times) radical 2</p>
<p>take half of that, and get the equation (m rad 2)/2 + h^2 = e^2</p>
<p>Then plug in m for e, and you will get that h equals M over rad 2.</p>
<p>I just did it, it works.</p>
<p>succinctly: I did (m/2)^2 + (m/2)^2 + h^2 = m^2 (cause m=e)</p>
<p>A is the answer if you play with the equation similar to the above.</p>
<p>thanks guys i did finally figure it out</p>
<p>To put it simply, use pythagorean theorem, and know that 45:45:90 degrees = 1:1: route2. That's all you need for this problem.</p>
<p>You need a strong application of the pythagorean theorem for a solid object. Then you have to figure out the ratio between m and h and...</p>
<p>I guess I'm not too good at explaining things.</p>
<p>You don't have to, just plug in #s and all your problems are solved! You can do it both ways, but you might make a mistake in doing it algebraically, and there are wrong answers that are waiting in the darkness for you to make a mistake(when you do it that way). I would use the algebraic way to check over my work if I wanted to make sure that I got it right.</p>
<p>Do questions like this really appear on the SAT?</p>
<p>Hey, I remember that problem. That was the one that I got right.:D</p>
<p>No, seriously, do a lot of problems like this come?</p>
<p>seriously, not a lot like this.</p>
<p>If you are unlucky, the most problems like this you will see on the SAT is 1...</p>
<p>They almost never have 3-D shape stuff.</p>
<p>Oh goodie. 'Cause I'd take ages to do something like this...</p>
<p>Whew! I took the SAT's in Oct. I guess I got lucky and dodged the bullet, because I somehow avoided this question.</p>
