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Find area: Given 3 coordinates of a triangles

ajsz033ajsz033 Posts: 169Registered User Junior Member
edited August 2009 in SAT Preparation
In the xy-coordinate plane, three vertices of a triangle are (1,2), (7,5), and (7,9). What is the area of the figure?

This is a grid in question. I had to use the distance formula to find the distance of all three sides of the triangle and then use Heron's formula to find the area.

S=(1/2)(A+B+C)
A=[S(S-A)(S-B)(S-C)]^(1/2)

This problem took me about 5 minutes because my method is so tedious and I easily make errors. Can anyone tell me if there is a shortcut to the problem?
Post edited by ajsz033 on

Replies to: Find area: Given 3 coordinates of a triangles

  • 12537291253729 Posts: 1,026Registered User Senior Member
    Was this problem from the College Board? Or is it from like Kaplan, Princeton Review...
  • YAHAYAHA Posts: 359Registered User Member
    In fact, there is.

    Took about a minute to do using my method. Basically you should sketch out the triangle with points and coordinates, but do not really follow any scale, just mark coordinates. Then use the following formula: S=1/2 base*height to find area of triangle. Also, you will have to introduce another point (7,2) and first find the area of the triangle with points (1,2),(7,2),(7,9). The area of this large triangle is S=1/2*(7-1)*(9-2)=21. Now find the area of triangle with points (1,2),(7,5),(7,2) using the same method. S=1/2*(7-1)*(5-2)=9. The answer to the original question is the latter subtracted from the former, i.e., 21-9= 12. Is this the correct answer?
  • jamesfordjamesford Posts: 3,447Registered User Senior Member
    Here's the simplest, most efficient method

    Plot the three coordinates just to get a visual of what you're solving for. The base of the triangle is easy to find since (7, 9) is just 4 vertical units above (7, 5). The height is the distance from (1, 2) to the line x = 7 (imagine a dotted line going from (1, 2) to (7,2) ). Therefore, height = 7 -1 = 6.

    Area = 1/2 (base)(height)
    = 1/2 (4)(6)
    = 12
  • YAHAYAHA Posts: 359Registered User Member
    Lol. You are right. Dont have to do all the extra calculations.
  • pokratpokrat Posts: 92Registered User Junior Member
    In my opinion , the simplest method to this type of problem would be to use the area of a triangle matrix determinant formula. You can google this method, but you might need to have some background in matrices.
  • bigbuddy93bigbuddy93 Posts: 63Registered User Junior Member
    here's the easiest way (in my opinion w/o using the calculations and crap) Convert the triangle into a Rectangle by just drawing a rectangle around the triangle (the vertices of triangle should be ON (not within) the rectangle. You will get 4 triangles (three of them will be right triangles) Find their formula and subtract it from the rectangle's formula.... you have the triangles area. Believe me it sounds crazy and all complicated but try it out. PM Me if you still don't get it!
  • YAHAYAHA Posts: 359Registered User Member
    ^ I think its in fact the longest method specified thus far. Even longer the the Heron's formula. Jamesford has specified the quickest and easiest method. You simply cannot get any more straightforward - just one half of base * height.
  • spratleyjspratleyj Posts: 582Registered User Member
    JamesFord's method is the correct one... I've never seen CB given this kind of problem without it being a right triangle....
  • cjgonecjgone Posts: 1,520- Senior Member
    Whenever you see a triangle problem like this, draw it out: the triangle will never be complex. I don't think the SAT has any math questions on it that require memorizing Hero's formula. You can easily solve this question because 2 of the points lie on the same x-position. Jame's method written above is the best.
  • xmexxmex Posts: 33Registered User Junior Member
    Lets call the points A B C

    Find the vector going from A to B, we'll call it V
    Find the vector going from A to B, we'll call it U

    Find the cross product of V and U

    Find the length of the cross product and divide by 2 and DONE!

    yeah seems like a long process but it can be done really quick on a calculator
  • spratleyjspratleyj Posts: 582Registered User Member
    You really shouldn't use your calculator on this... just sketch it out and the rest is easy... takes like 10 seconds....
  • Sci-FrySci-Fry Posts: 1,316Registered User Senior Member
    This took me about 30 seconds in my head - there's one very simple thing that sticks out in the question that makes it faster to do.

    So we know that the area of a triangle is base x height/2. Notice that two of the points on the triangle have a common x-coordinate (7). So we can use these coordinates to form the base. The length of the base is 9 - 5 = 4, since that is the difference between the y-coordinates of the two points.

    The next thing is to find the height, which is simply 7 minus the x co-ordinate of the third point - i.e. 7 - 1 = 6.

    Now we do 6 * 4/2 = 12.

    It might be worth imagining the triangle in your head or doing a rough sketch. Also note that I kind of turned the graph around to form the base and height of the triangle.
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