gameguard

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Well, I was always pretty good at math and didnt think it to be much of a problem. However, ive been doing the topscore problems and some of the questions they give are not very "standard math" ... at least in my book.

There was numerous problems involving very random equations (given in the solutions). For example, there was a question with 3 coordiantes and I was supose to find the area of the triangle - which was not any special triangle. the eq. they gave was :
A = 1/2[(x1y2 + x2y3 + x3y1) - (y1x2 + y2x3 + y3x1)]

I duno about you guys but I have never in my life come across this equation. Im guessing its some variation of herons formula and distance formula...

I mean I could use distance formula for each side and then plug it in to herons formula, but that would be VERY messy and time consuming. This is just an example of the kind of things that caught me off guard. There was alot of questions that sported unorthodox equations in their solutions (eqs that wernt covered in kaplan and are even hard to come by on the net)

I guess im rambling... what im trying to get at is this. For those of you who took the test, how many times did you come across a question that required you to know some kind of special formula to solve in any reasonable amount of time?

Also, if you guys could add some equations that you found useful I would really appreciate it.

Thank you~
 

wclubin

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gameguard said:
Well, I was always pretty good at math and didnt think it to be much of a problem. However, ive been doing the topscore problems and some of the questions they give are not very "standard math" ... at least in my book.

There was numerous problems involving very random equations (given in the solutions). For example, there was a question with 3 coordiantes and I was supose to find the area of the triangle - which was not any special triangle. the eq. they gave was :
A = 1/2[(x1y2 + x2y3 + x3y1) - (y1x2 + y2x3 + y3x1)]

I duno about you guys but I have never in my life come across this equation. Im guessing its some variation of herons formula and distance formula...

I mean I could use distance formula for each side and then plug it in to herons formula, but that would be VERY messy and time consuming. This is just an example of the kind of things that caught me off guard. There was alot of questions that sported unorthodox equations in their solutions (eqs that wernt covered in kaplan and are even hard to come by on the net)

I guess im rambling... what im trying to get at is this. For those of you who took the test, how many times did you come across a question that required you to know some kind of special formula to solve in any reasonable amount of time?

Also, if you guys could add some equations that you found useful I would really appreciate it.

Thank you~

Well the above equation clearly comes from the standard formula for the equation of a triangle namely Area equals 1/2 the base times the altitude. Both the base and the altitude used the distance formula and so that is why there are no square root signs left. when you mult a square root by another square root the square roots go away. And so there for you are left with !/2(stuff)(stuff). after squaring everthing out in the inside and bringing all neg terms to one side and factoring out the negative sign you arive at what you get above. It is just the standard formula with some algebraic manipulation. no new formulas. I would have to see the specific question to answer more specificly.
 
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gameguard

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uhh... well the question is :

In a two dimensional coordinate system, point A=(3,-6), point B=(1,1) and point C=(7,6). Find the area of the triangle ABC.

Answer is 26



I still cant seem to come up with the formula given ... This triangle is not any kind of standard triangle (at least i dont think) so .5bh seems pretty complicated.


also any aditional help regarding my rant up there would be appreciated :0
 

sonysonysony

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Whenever you have a (x, y) point system for the 3 points of the triangle, you will need to find the distance between each of the two points. To find the distance b/w the points (x1, y1) and the points (x2, y2), this equation must be used.

(d)square=(x1-x2)square+(y1-y2)square

I think they just combined this equation with 1/2bh to get the equation in that book.

Hope this helps.

later


gameguard said:
uhh... well the question is :

In a two dimensional coordinate system, point A=(3,-6), point B=(1,1) and point C=(7,6). Find the area of the triangle ABC.

Answer is 26



I still cant seem to come up with the formula given ... This triangle is not any kind of standard triangle (at least i dont think) so .5bh seems pretty complicated.


also any aditional help regarding my rant up there would be appreciated :0
 

DMD to Be

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There must be an easier way to work this problem...Besides, you have less than a minute to do it...no time to do 3 distance formulas and the 1/2 bh...
 
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gameguard

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really... and how can you find the height when its not a right triangle and there are no angles.
 

luder98

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I saw a question similar but easier on my DAT. The one I had was a special case when two points are on one horizontal line. Here is how I did it:
- Draw a small xOy plane
- Put all the three points into the plane
You can quickly find one base based on the two points on the same horizontal line (== the absolute value of the difference between the x-values). The height is the absolute value of the difference between the third y-value with the y-value of the other two. Then use 1/2bh formula.
You can use this method in similar manner if two of the three points are on one vertical line.
For your question, the quickest way (if you don't remember the formula) is to find one base from two points. Find the equation pass through these two points. Find the distance from the third point to this equation. Then use 1/2bh formula. If you can do all that in less than 1 min, then do it. Otherwise, just mark and move one. There are many more much less calculating questions later on. You don't want to miss those. Good Luck!
 
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gameguard

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luder98 said:
I saw a question similar but easier on my DAT. The one I had was a special case when two points are on one horizontal line. Here is how I did it:
- Draw a small xOy plane
- Put all the three points into the plane
You can quickly find one base based on the two points on the same horizontal line (== the absolute value of the difference between the x-values). The height is the absolute value of the difference between the third y-value with the y-value of the other two. Then use 1/2bh formula.
You can use this method in similar manner if two of the three points are on one vertical line.
For your question, the quickest way (if you don't remember the formula) is to find one base from two points. Find the equation pass through these two points. Find the distance from the third point to this equation. Then use 1/2bh formula. If you can do all that in less than 1 min, then do it. Otherwise, just mark and move one. There are many more much less calculating questions later on. You don't want to miss those. Good Luck!

yea, i got ya. Thanks for the reply~
 

DrJimLa

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was very interested by the ? bc i saw one similiar on my practice test....
if you have three coordinates and one is the origin (or you can move all points to make one the origin the equation is fairly easy)
so for your example
A (1,1)-------------becomes (0,0)
B (3,-6)-----------------------(2,-7)
C (7,6)------------------------(6,5)
then use the equations
Area= 1/2 [(Bx*Cy)-(By*Cx)] and those are absolute value brackets
Area= 1/2 [2*5-(-7*6)]
A= 1/2 [10+42]= 1/2 [52]
A=26
hope thats is useful and it shows up on DAT for someone! :thumbup:
 

tom_servo_dds

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DrJimLa said:
was very interested by the ? bc i saw one similiar on my practice test....
if you have three coordinates and one is the origin (or you can move all points to make one the origin the equation is fairly easy)
so for your example
A (1,1)-------------becomes (0,0)
B (3,-6)-----------------------(2,-7)
C (7,6)------------------------(6,5)
then use the equations
Area= 1/2 [(Bx*Cy)-(By*Cx)] and those are absolute value brackets
Area= 1/2 [2*5-(-7*6)]
A= 1/2 [10+42]= 1/2 [52]
A=26
hope thats is useful and it shows up on DAT for someone! :thumbup:
Great help, thanks!
 

dental#1

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This problem is on Topscore and my answer was real one I think,so let me know!I think You need to carry the negative sign. I am meeting with a tutor today so I will be able to see if there is an easier way

(X1Y2 + X2Y3 + X3Y1) - (Y1X2 + Y2X3 + Y3X1)

(3*1 + 1*6 + 7*-6) - ( -6*1 + 1*7 + 6*3)

(3*1 + 1*6 + 7*-6) - ( 6*1 - 1*7 - 6*3)

3+6+ (-42) + 6 -7 -18

-33-19= -52 1/2 = 26
 
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