so i don't know if everyone noticed...

Test Preparation ACT Preparation
1

<p>was math section so easy or what? lol
It was my 7th or 6th times taking the act, but i think that this math section was the easiest except the question about 3 cups of juice concentraion and 5 cups of water thing....</p>

2

<p>I thought that this test was easy in its entirety, with the exception of the reading section. I liked the math portion because, as you said, it seemed easy. I was quite confident with the majority of my answers, and then even more confident once I found out others got the same. I did miss the area of a rhombus, which was an easy one. Too bad geometry was freshman year and I didn’t remember that. >.< Besides that mishap, the test went well. The last ten or so questions were exceedingly easy relative to how they normally are… or I’ve just been practicing with Barron’s too much.</p>

3

<p>so what was the rhombus formula again?? crap, I hope i didn’t get that wrong. I felt like the questions were getting easier and easier as I was reaching the end. wouldn’t you agree?</p>

4

<p>In a way I agree, it seemed that the middle questions were harder than the end ones, and the middle questions weren’t terrible hard anyways. </p>

<p>The area is one half of the product of the diagonals. One diagonal was 6 and the other was 8, so it was .5(6)(8) which is 24. Seeing that the diagonals were 3+3 and 4+4 respectively, I assumed that they hypotenuse must be 5. Then, I multiplied the base hypotenuse and a side hypotenuse to get 25. What a shame.</p>

5

<p>awwwww good! I was guessing between 24 and 48 on that question and i guessed it right lol.</p>

<p>Anymore hard questions? I honestly don’t remember the hard questions except that weird one. So… what do you think the curve will be for the math section?? Harsh?</p>

6

<p>I don’t really remember any other hard questions. That one wasn’t really hard either, I just didn’t remember the formula. :confused: If I’d have thought it through, I could have separated it into four different triangles and found the sum of their areas, too bad I didn’t think of that until I realized that I got it wrong. :P</p>

<p>I’m sure that the curve will be harsh; but I’d rather have an easy test with a harsh curve than a hard test with a generous curve any day.</p>

7

<p>There was an rea of a rhombus question? Can you reiterate the problem please?</p>

8

<p>It was insignificant if you knew the area formula for it already. It gave us a rhombus formula and showed that one diagonal was 8 and the other was 6, and then asked for the area. The correct answer was 24.</p>

9

<p>Oh, that one. I didn’t remember it at first because I just found the area of the top triangle, then multiplied that area by 2. Thanks for the reminder, though.</p>

10

<p>was 25 an option for the rhombus one? ( i know the correct answer was 24. I’m just asking because if 25 was an option I probably put that)</p>

11

<p>can’t you just use base x height for a parallelogram?</p>

12

<p>25 was an answer for the rhombus one, because I remember picking it with somewhat confidence that I was right. I thought that the base X height would work for it, but you can’t find the base because there isn’t sufficient information to determine it. (Angles aren’t right so you can’t use Pythagorean’s formula nor could you use trigonometry.)</p>

13

<p>I feel like I don’t remember this picture… but for some reason I felt like i just multiplied 2 numbers and got 24… hmm…</p>

<p><a href=“http://img585.imageshack.us/img585/9343/75389350.jpg[/url]”>http://img585.imageshack.us/img585/9343/75389350.jpg&lt;/a&gt;&lt;/p&gt;

<p>so it wasn’t the top one? It was the bottom one?</p>

14

<p>^Yes.</p>

<p>For the all the people who didn’t know the formula, you could have derived it from a square.</p>

<p>Area of square with side length ‘a’ = a^2
length of diagonal= sqrt(2a^2) = sqrt(2)*a</p>

<p>So how do you get from sqrt(2)* to a^2?</p>

<p>you multiply by the other diagonal and divide by 2:</p>

<p>([sqrt(2)* a] * [sqrt(2)*a])/2 = a^2 = area of square</p>

<p>So the general formula for a quadrilateral with congruent sides (implies perpendicular intersections at diagonals) = (d1*d2)/2</p>

<p>At least that’s what I did. </p>

<p>Or…you could have remembered the formula, that works too. ;)</p>