#$&* course MTH 151 02/27/2014 at 9:31 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: Each corner of the square will connected to each of the other three corners, so from each corner you would have drawn three lines. Since there are four corners, had you followed the instructions precisely you would have drawn 4 * 3 = 12 lines. However each of these lines will be identical with another line you would have drawn, since for any two corners you would be drawing a line from the first to the second then another overlapping line from the second to the first. Therefore you might have said that there are 6 lines. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I do understand why there would be twelve lines drawn, but you end up with six lines.
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Given Solution: You should have a total of 8 triangles. The diagonals divide the square up into 4 small triangles. Each diagonal also divides the square into 2 larger triangles. Since there are 2 diagonals there are 4 larger triangles. The 4 small triangles and the 4 larger triangles total 8 triangles. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. The lines you drew should form a square with its two diagonals. Label the corners of the square A, B, C and D, going in order around the square, and label the center where the diagonals cross E. Now list all possible combinations of 3 of the letters A, B, C, D, E (note: combinations don't care about order, so A D E is the same as D A E or E A D or any other combination of these same three letters, so list each possible combination only once. That is, if you list for example ADE you won't list DAE). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There will be ten combinations: ABC, ABD, ABE, ACE, ACD, ADE, BCD, BCE, BDE, and CDE. confidence rating #$&*: 3. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The easiest way to list these sequences is alphabetically: ABC, ABD, ABE all start with AB; then ACD and ACE start with AC and ADE starts with AD. This is a list of all possible combinations containing A. We next list all possible remaining combinations containing B: BCD, BCE and BDE. Then we write down CDE, the only remaining combination containing C. We thus have the 10 combinations ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. Of the 10 combinations ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE, which form triangles on your figure? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ABC, ABD, ABE, ACD, ADE, BCD, BCE, CDE all form triangles. confidence rating #$&*: 3. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ABC forms a large triangle from A to B to C and back to A. The same is true of ABD, ACD and BCD. These are the large triangles in the figure. ACE and BDE form straight lines, not triangles. ABE, ADE, BCE and CDE form small triangles. Thus of the 10 possible combinations of labeled points, we find the 4 large triangles and the 4 small triangles we saw earlier, in addition to 2 straight lines which do not for triangles. Since any triangle in the figure must be labeled by three of the five points A, B, C, D, E, we see that these are the only triangles that can be formed. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. Sketch the same figure as before, but without the line segment from A to B. Now how may triangles are there? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: There will be five triangles: ADE, DEC, BCE, BCD, ACD. confidence rating #$&*: 3. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Any possible triangle must still come from the list ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE. We again see that ACE and BDE form straight lines so do not count as triangles. Now ABC, ABD and ABE do not form triangles because the line segment AB is now missing. This leaves us the five triangles ACD, ADE, BCE, BCE and CDE. ********************************************* Question: `q006. How many triangles in the original figure do not have point C as a vertex? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If ABC, ABD, ABE, ACD, ADE, BCD, BCE, CDE form triangles, all the combinations with C are excluded. So, three triangles: ABD, ABE, ADE do not have C as a vertex. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating:" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!