<p>triangle has vertices at (4,5) (5,3) (1,3) respectively. What is the area?
a. 4
b. 4 sqrt(2)
c. 4 sqrt(3)
d. 8
e. 8 sqrt(2)</p>
<p>Draw a graph</p>
<p>|(x1y2-x2y1)+(x2y3-x3y2)+(x3y1-x1y3)|/2
You just have to use trapezoids to derive the formula
Answer is a. 4</p>
<p>Just imagine a graph or subtract the x and y values.<br>
The height is 2. The base is 4.</p>
<p>Your answer is 4</p>
<p>The side with (5,3) and (1,3) is parallel to the x-axis.</p>
<p>Area = (1/2)(4)(2) = 4.</p>
<p>@xisheng, an easier way to derive the area of a triangle in general is to take the magnitude of the cross product of two vectors formed by two of the sides, then multiply by 1/2.</p>
<p>Good grief! The easiest thing is to draw the triangle and realize that the base is 4 and the height is 2. Vectors? Really?</p>
<p>@rspecnce You could use vectors, but the process is superfluous. The most direct method is to use the formula given, since it directly corresponds with the information given. But I do agree with you that there are many options to solve the problem. There are many shortcuts depending on the problem, but for people that aren’t mathematically inclined they might prefer to approach problems formulaically, thus saving them time and trouble of over thinking the problem.</p>
<p>Sure, I guess, although it’d take me a bit of memory to remember that formula…then again, all the variables can be permuted to get the correct formula.</p>
<p>another way to solve this problem would be with matrices, though you may not have learned this method. presuming you have:</p>
<p>[4 5 1]
[5 3 1]
[1 3 1]</p>
<p>By evaluating the determinant, you get 12+5+15-(3+25+12) = -8
-8*1/2= absolute value(-4) = 4 Ans.</p>
<p>lol why would you use vectors and matrices for this hahaha, you should keep this simple</p>
<p>We’re just saying, in general, if none of the sides are parallel to an axis (or if we’re working in 3-space), it might be easier to use cross products.</p>
<p>Re post #3:
An alternate notation might be easier to memorize:</p>
<p>|x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|/2.</p>
<p>@gcf101
I don’t think anyone would find it easier to memorize that formula. The determinant formula is very simple, it’s just 1/2 * [one column of x-coordinates, one column of y-coordinates, one column of 1s]. (That’s how I remember it anyway).</p>
<p>^Anyone who is not familiar with the 3x3 Det would.</p>
<p>Never take vectors in the ACT !!!</p>
<p>^I thought ACT tests up to pre-calc…vectors are part of the standard pre-calc curriculum. </p>
<p>Besides, vectors are awesome.</p>
<p>I’m in honors linear algebra 2 right now at university and I think that’s a completely ridiculous way to approach the question given that it’s such a simple area question to begin with.</p>
<p>Well, we were mostly just talking about how to find the area of a triangle given <em>any</em> three points in space. For this particular ACT problem the base is parallel to the x-axis so it’s easy.</p>