<p>In the figure above, AB=CD. What is the value of t?</p>
<p>That's the link to the figure. It's on the SAT Blue Book second edition page 397 section 3 question #6</p>
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<p>Okay you know that the two line segments are equal, so the the lengths are equal as well. This means that the distance from A to B is the same as from C to D. The distance from C to D is 9, so the distance from A to B must be 9 as well. THis means t has to be -6 for the distance from A to B to equal 9. The answer is C.</p>
<p>thanks. But you mean 10 points. The answer is C, but it’s -7</p>
<p>No, he means 9 points. 5 minus -4 = 9.</p>
<p>The answer is C, -6. A simple and quick way to solve this question is to compare the vertical distance of AB to the horizontal distance of CD which as the question states, are equal. This way you don’t even need to use the distance formula. Since the x-distance on CD is from -4 to 5, the distance is nine. Now look at the y-distance of AB. It’s already 3 spaces above the x-axis, so for it to have a distance of 9 it needs to have a length of 6 spaces below the x-axis. This leads to answer C, or -6.</p>
<p>Hi, since you mentioned the distance formula: I tried to work it out using the distance formula, but I got answers like 12 and -12. Can anyone demonstrate the solution using the distance formula?</p>
<p>^
Okay… instead of memorizing “distance formulas”, start to understand what exactly you’re doing. The SAT is trying to nail you when it comes to going by the rules and not really understanding your math.</p>
<p>The distance is simply the hypotenuse of a right triangle. Since the line is perfectly horizontal, there is no need in doing that; you can calculate its distance using the difference in x-coordinates.</p>
<p>To explain your confusion, though:</p>
<p>a^2+b^2=c^2. Let a be delta x and b be delta y. b = 0 so a=c=9.</p>
<p>^ Way more complicated, just use distance formula, which is derived from Pythagorean theorem (same thing but simpler)</p>
<p>^ What? 0_0 I didn’t suggest any more complicated method, just demonstrated how the “distance formula” is ineffective with lines set on a scale, since a^2 = c^2… And that formula is not derived from the Pythagorean theorem, nor is it simpler. It’s the exact same thing.</p>
<p>Could anyone explain question 19 of the same section?</p>
<p>The pyramid question.</p>
<p>This is what I did, tell me where I went wrong.</p>
<p>m^2=h^2+(e/2)^2
m^2=h^2+e^2/4
(e=m)
e^2=h^2+e^2/4
h^2=3e^2/4
h=Root of 3*e/2</p>
<p>So my answer is A, which is wrong.</p>
<p>^Use half of the diameter of the base square in the Pythagorean theorem.</p>
<p>That’s what I did. e/2.</p>
<p>Check your figure again. e/2 is half of one side of the base. Check the point where h meets the base and evaluate the distance to one of the corners of the base. The distance is 1/2 of the diagonals of that base, not 1/2 of the sides.</p>
<p>^Which is eSQRT(2)</p>
<p>Yep, try to find the length of half the DIAGONAL of the square base. It turns out to be e/(2^1/2). Use this length to determine h in terms of e.</p>
<p>The mindless mistakes you make… Argh. Thanks everyone.</p>
<p>I made a thread about it, but got no replies. I hope I get some input here… 
I was doing Barron’s 2400’s math section. I was shocked to find conditional probability in it! I have never encountered a question pertaining to conditional probability in the SAT or its practice tests! Probability is in the syllabus for sure, but conditional probability? </p>
<p>And is simultaneous linear inequality, wherein we have to solve two inequalities for x and y in our course for SAT? (In which we have to plot the equations, and the highlight the areas, and then see the overlapping area.)</p>