<p>I need to find the area of triangle ABC that has the coordinates A(1, 3, 0), B(0, 2, 5), and C(-1, 0, 2). This prob. is killing me!</p>
<p>Also: (if you can)</p>
<p>What is the sum of all the pos. digit integers that can be formed from the digits 2, 3, 5, 6, and 7? Note that the same digit can appear more than once.</p>
<p>Ok, for your first question...</p>
<p>If you are familiar with vector math, you can use the fact that 1/2 the cross product of two vectors that make up a triangle gives the area of the triangle. Since you are given three distinct points, making two vectors is easy, as you just have to subtract the coordinates.</p>
<p>If you want a clearer explanation of this process, please reply.</p>
<p>For question number two....</p>
<p>Im not sure I understand the question as it is presented. More specifically, what do you mean by:
" What is the sum of all the pos. digit integers " -- what is meant by digit integers, are you missing the number of digits?</p>
<p>For question 1:</p>
<p>I'm in Algebra 2 and the question is supposed to be geared towards ALgebra 2 students....so vectors are out... is there a simpler solution? </p>
<p>Question 2:
That's how the question was written! I know what you mean! I said to myself the same thing! I had no idea what it exactly meant!</p>
<p>Anyway, if you can help me with Q. 1, I would be fine with that....</p>
<p>Oh, thanks by the way...</p>
<p>ya sat I math isn't this complicated..</p>
<p>would the distance distance formula work?</p>
<p>The problem can be done with only algebra, but its kind of messy. Anyways, heres how you'd go about doing it.</p>
<p>Using the distance formula, the length of the sides of the triangle can be calculated. From this, we find AC = sqrt(17), AB= sqrt(27), BC = sqrt(14).</p>
<p>The area of a triangle is 1/2bh. So let's pick side AC arbitratily as the base, so we have b = sqrt(17). Now how do we find h, the height?</p>
<p>I would advise drawing the triangle on paper. Draw a line from B to AC, our base, such that B is perpindicular to AC. Lable the point at which this line intersects AC as "P". This line is the height of the triangle. You will notice that this line splits the triangle into 2 right triangles, CBP and PCA. </p>
<p>Now, denote the length of CP as x, and BP as h. Using the pythagorrean theorem, we can set up the following two equations pertaining to these newly formed right triangles:
x^2 + h^2 = BC^2 = 14
(sqrt(17)- x)^2 + h^2 = AC^2 = 27. </p>
<p>Thus, we have a system of equations that we can solve for x and h.Once you solve for h, you can multiply 1/2bh to get the area.</p>
<p>I find it hard to believe a problem of this length would be on the SAT, unless I am missing something very obvious.</p>
<p>what is "sqrt(17)?" what is sqrt?</p>
<p>short hand for square root</p>
<p>Thank you so much adimeola!</p>
<p>Well, let's go Algebra 2 way.</p>
<ol>
<li>x-coordinates of points A( 1,3,0) and C(-1,0,2) are opposite numbers.</li>
<li>Points A and C are equidistant from plane yz.</li>
<li>Midpoint M of AC is in plane yz.</li>
<li>Coordinates of M:
(1,3,0) + (-1,0,2) ) / 2 = (0,3/2,1).
See footnotes 4.1 and 4.2.</li>
<li>h is a height of triangle ABC from A to BM.</li>
<li>Area of triangle AMB:
A(AMB) = 1/2 (BM) h.</li>
<li>BM is median of triangle ABC.
A(AMB) = A(CMB),
A(ABC) = (2) A(AMB),
A(ABC) = (BM) h.</li>
<li>Distance between B(0,2,5) and M(0,3/2,1)
BM = sqrt(65) / 2 .</li>
<li>Since line BM is in yz-plane and goes through points B(2,5) and M(1/2,1),
its equation is z = (8)y - 11, or
(8)y + (-1)z + (-11) = 0.
(x=0 for all points of line BM).
Equation of line BM in xyz-space:
ax + by + cz + d = 0, or
(0)x + (8)y + (-1)z + (-11) = 0, </li>
<li>h is a distance between point A(1,3,0) and line BM.
Formula:
h = | a(1) + b(3) + c(0) + d | / sqrt(a^2 + b^2 + c^2)
h = | (0)(1) + (8)(3) + (-10)(0) + (-11) | / sqrt (0^2 + 8^2 + (-1)^2)
h = 13 / sqrt(65)</li>
<li>A(ABC) = (BM) h.
A(ABC) = (sqrt(65) / 2 ) * ( 13 / sqrt(65) ) = 13 / 2 = 6.5</li>
</ol>
<p>6.5 !!!!!!!!
And all the work :(
It's gotta be some much shorter solution.
This one was just the most direct that came to my head.</p>
<h2>I also hope I did not make some dumb mistakes.</h2>
<p>Footnotes.
4.1 Instead of wading through steps 1-3 we could just find coordinates of midpoint M and notice that x-coordinate of M(0,3/2,1) is 0.<br>
That means point M is in plane yz.
4.2 On the other hand, even step 1 might not be obvious.
I strongly suggest you carefully draw 3-D diagram of triangle ABC in xyz coordinate system.
If you look at AC projection into xy-plane (say, AD), and AD y-intercept (say, N), you may notice that AN=ND, so yz-plane "cuts" AC in half, and AC midpoint M is in yz-plane (M projection into xy-plane is N).
Sounds scary, but drawing this could turn to be a lot of fun.</p>
<p>
[quote]
all the pos. digit integers
[/quote]
</p>
<p>It's probably n-digit integers: 3-digit, or 4-digit, or...</p>
<p>I'd go along with gcf101. I think what the question meant to say was:</p>
<p>'How many k-digit integers can you form using the pool of digits 2,3,5,6,7? '</p>
<p>I would suspect they are looking for a derived formula f(k). For k=1, it's easy; f(1) = the # different digits in the pool = 5. For k=2, there are (5)(5) = 25 possibiliities, from 22,23... all the way to ... 76,77 . In general, f(k) = 5 ^ k .</p>
<p>The easiest way to do this would be to use Heron's theorem.
Suppose you have a triangle with lengths A=3, B=4, and C=5. You can probably calculate that the area of this triangle would be equal to 6. However, using Heron's theorem, we find the perimeter of all the sides and divide by two, designating the number we get as S. So, in this case, S= (3+4+5)\2=6. The area of the triangle can be calculated by finding the square root of (S)(S-A)(S-B)(S-C) or in this case the square root of 6<em>3</em>2*1. This would give us the square root of 36, which simplifies to give the area of the triangle as 6.</p>
<p>This works for ANY Triangle- meaning you can now find the areas of triangles that do not have right angles!!!!</p>
<p>If you use Heron's formula, doesn't the area come out to be 7.65?</p>
<p>AB = sqrt(27)
AC = sqrt(17)
BC = sqrt(14)</p>
<p>Now S = (sqrt(27)+sqrt(17)+sqrt(14))/2 = approx 6.53</p>
<p>Now sqrt((6.53)(6.53-sqrt27)(6.53-sqrt17)(6.53-sqrt14)) = 7.65</p>
<p>What am I doing wrong?</p>
<p>My bad. I forgot- Haren's formula can be used only when the sides are integers.</p>
<p>got damn wut kind of math is this precalculus</p>
<p>I wouldn't know.... I skipped Precalculus</p>
<p>I was thinking exactly what adimeola was thinking with the vectors. It's kinda funny how much easier that makes it.</p>
<p>Could you explain how to do it with vectors? I dont really recall them.</p>