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I disagree with your logic, the question is how many triangles can be formed with all the available points of intersection, not how many arrangements are possible to form one triangle. Your solution is for the latter.prindaroy wrote:The answer is 10. 5c3 = 10
5 lines, to form a triangle, you need 3 lines, so how many ways can you choose 3 out 5?
5c3 = 10
Actually, with four lines you will have 6 different points of intersection and with 5 lines you'll have 10. Draw the whole thing to double check. This is why your reasoning is flawed.prindaroy wrote:With 3 lines, you can form 3 distinct points of intersections. With four lines you can form four distinct points of intersection. So for 5, you can point five distinct points of intersection. So think about it this way. Let's say with five points of intersection, you have a pentagon. How many triangles can you make out of a pentagon? You can make 10 triangles because 5c3. I know my logic is correct because this is a kaplan question which I answered a few days ago.
prindaroy wrote:With 3 lines, you can form 3 distinct points of intersections. With four lines you can form four distinct points of intersection. So for 5, you can point five distinct points of intersection. So think about it this way. Let's say with five points of intersection, you have a pentagon. How many triangles can you make out of a pentagon? You can make 10 triangles because 5c3. I know my logic is correct because this is a kaplan question which I answered a few days ago.
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