» CC HOME » FORUM HOME

 College Confidential Find area: Given 3 coordinates of a triangles
 User Name Remember Me? Password New User

Welcome to College Confidential!
The leading college-bound community on the web
Join for FREE now, and start talking with other members, weighing in on community polls, and more.

Also, by registering and logging in you'll see fewer ads and pesky welcome messages (like this one)!
 Discussion Menu »Discussion Home »Help & Rules »Latest Posts »NEW! CampusVibe™ »Stats Profiles Top Forums »College Chances »College Search »College Admissions »Financial Aid »SAT/ACT »Parents »Colleges »Ivy League Main CC Site »College Confidential »College Search »College Admissions »Paying for College Sponsors SuperMatch - The Future of College Search! CampusVibe - Almost As Good As A Campus Visit!
 08-12-2009, 01:45 AM #1 Junior Member   Join Date: Apr 2009 Posts: 163 Find area: Given 3 coordinates of a triangles 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? Reply
 08-12-2009, 01:56 AM #2 Member   Join Date: Jul 2009 Location: Connecticut--->UPenn Posts: 975 Was this problem from the College Board? Or is it from like Kaplan, Princeton Review... Reply
 08-12-2009, 02:04 AM #3 Member   Join Date: Jun 2009 Posts: 344 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? Reply
 08-12-2009, 02:22 AM #4 Senior Member   Join Date: Jan 2008 Location: West Coast ---> MIT '14 Posts: 2,318 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 Reply
 08-12-2009, 02:31 AM #5 Member   Join Date: Jun 2009 Posts: 344 Lol. You are right. Dont have to do all the extra calculations. Reply
 08-12-2009, 04:53 AM #6 Junior Member   Join Date: Aug 2009 Posts: 90 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. Reply
 08-12-2009, 03:10 PM #7 Junior Member   Join Date: Dec 2008 Posts: 63 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! Reply
 08-12-2009, 04:16 PM #8 Member   Join Date: Jun 2009 Posts: 344 ^ 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. Reply
 08-12-2009, 04:41 PM #9 Member   Join Date: Jan 2009 Posts: 572 JamesFord's method is the correct one... I've never seen CB given this kind of problem without it being a right triangle.... Reply
 08-12-2009, 06:52 PM #10 Senior Member   Join Date: Jun 2009 Location: Washington -> UC Berkeley '15 Posts: 1,200 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. Reply
 08-12-2009, 10:51 PM #11 Junior Member   Join Date: May 2009 Posts: 33 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 Reply
 08-13-2009, 12:58 AM #12 Member   Join Date: Jan 2009 Posts: 572 You really shouldn't use your calculator on this... just sketch it out and the rest is easy... takes like 10 seconds.... Reply
 08-13-2009, 01:30 PM #13 Senior Member   Join Date: Oct 2007 Location: Italy --> Oxford '14 Posts: 1,216 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. Reply

 Bookmarks