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| Math Problem | |
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| Tweet Topic Started: Mar 8 2006, 04:07 PM (464 Views) | |
| dolmansaxlil | Mar 8 2006, 04:07 PM Post #1 |
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HOLY CARP!!!
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(excuse my awful MS Paint diagram) Angle x = 90 degrees What is the area of rectangle ABCD? (and how did you find it?) |
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"Your first 10,000 photographs are your worst." ~ Henri Cartier-Bresson My Flickr Photostream | |
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| CrashTest | Mar 8 2006, 04:08 PM Post #2 |
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Pisa-Carp
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Wow, you want to not be popular really fast, huh!? |
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| Aqua Letifer | Mar 8 2006, 04:12 PM Post #3 |
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ZOOOOOM!
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SIDE SIDE ANGLE. The ambiguous case. Law of sines, or cosines should do nicely. Heck, don't even have to do that. Pythag's theorem, and using the rectangle (if you wanted to), with it's complimentary angles, should be more than enough. |
| I cite irreconcilable differences. | |
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| dolmansaxlil | Mar 8 2006, 04:13 PM Post #4 |
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HOLY CARP!!!
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Ok. Now imagine you're in grade 8. |
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"Your first 10,000 photographs are your worst." ~ Henri Cartier-Bresson My Flickr Photostream | |
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| Aqua Letifer | Mar 8 2006, 04:14 PM Post #5 |
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ZOOOOOM!
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AHHHHH. Dunno if I could do that at age 8! And I'm real bad at explanations. I'll give it a go, though, it'll just take a bit. |
| I cite irreconcilable differences. | |
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| dolmansaxlil | Mar 8 2006, 04:15 PM Post #6 |
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HOLY CARP!!!
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Grade 8. So 13ish. They have just studied Pythagorean Theorem, and they are currently working on congruency. They know the formulas for area. |
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"Your first 10,000 photographs are your worst." ~ Henri Cartier-Bresson My Flickr Photostream | |
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| Nunatax | Mar 8 2006, 05:12 PM Post #7 |
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Middle Aged Carp
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An alternative way, should they not have had trigonometry yet (I'm assuming you switched the 6 and 8, as the 8 is shorter than the 6 in your drawing :crazy:) : ![]() [cb] = sqrt (8² + 6²) = 10 So, y = 10 - z 8² = y² + h² <=> 8² = (10-z)² + h² <=> h² = 8² - (10-z)² (equation 1) Also : 6² = z² + h² (equation 2) Now put equation 1 in equation 2 : 6² = z² + 8² - (10 – z)² <=> 36 = 64 + z² -100 + 20z – z² <=> 36 = -36 + 20z <=> 72 = 20z <=> z = 3.6 Put z in equation 1 : h = sqrt (64 – (10- 3.6)²) = sqrt (64 – 40.96) = sqrt (23.04) = 4.8 10 * 4.8 = 48 which is the area of the rectangle. |
| You seem somewhat familiar. Have I threatened you before? | |
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| dolmansaxlil | Mar 8 2006, 05:20 PM Post #8 |
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HOLY CARP!!!
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Nope. The algebra is too advanced. (This question was given on a test today. The teacher was out of the building and I was in helping my caseload kids. Though I could get the answer, in much the way you did, I was unable to get the answer using skills they know. My guess is that the diagram should have contained one other measurement - X to either a or b. But I'm trying to figure out a way you can do it using conguency and the pythagorean theorem only, without the more complex algebra) |
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"Your first 10,000 photographs are your worst." ~ Henri Cartier-Bresson My Flickr Photostream | |
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| Nunatax | Mar 8 2006, 05:36 PM Post #9 |
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Middle Aged Carp
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I think I got it Triangle bcx is congruent with the two triangles I made by drawing h. So 10/8 = 6/h, so 10 * h = 6 * 8 <=> h = 48/10 <=> h = 4.8 Do I hear you slapping your head too?
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| You seem somewhat familiar. Have I threatened you before? | |
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| Anonymous | Mar 8 2006, 05:37 PM Post #10 |
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Junior Carp
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No, but someone is going to get slapped. And I'm looking at you, big boy. |
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| pianojerome | Mar 8 2006, 05:40 PM Post #11 |
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HOLY CARP!!!
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48. a^2 + b^2 = c^2 36 + 64 = 100 c = 10 (that's the length of the bottom (and the top) of the rectangle) sin b = opposite / hypotenuse sin b = 6 / 10 b = 36.86 The corner is 90°, so the angle at b outside of the triangle is 90-36.86 = 53.13. Sum of angles in a triangle is 180: 180-90-53.13 = 36.86 sin 36.86 = s / 8 s = 4.8 4.8 x 10 = 48. Am I right? |
| Sam | |
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| mrenaud | Mar 8 2006, 05:46 PM Post #12 |
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Middle Aged Carp
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Might this one be simple enough? Basically, a triangle containing a 90-degree angle is half a rectangle, therefore you could multiply the length of the catheti and divide that by two (0.5*6*8 = 24). According to Pythagoras' Theorem, the hypotenuse would be sqrt(6^2+8^2) = 10. If area = 0.5 * base * height, then height = area / (0.5 * base), or 24/5 = 4.8. So we now have both sides of the rectangle, 10 * 4.8 = 48. |
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| big al | Mar 8 2006, 06:04 PM Post #13 |
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Bull-Carp
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Extending mrenaud's analysis, if you divide the triangle into two triangles by dropping a vertical from the right angle to the opposite side, you can see that you have two rectangles each with a diagonal. In other words, the remaining area is the area of two triangles above the diagonals. But you already know that the area of the original triangle is 24 so the total area must be simply twice that of the original triangle since the remaining pieces are equal to the two pieces of the original triangle. You don't need to invoke Pythagoras or any trig functions. Big Al |
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Location: Western PA "jesu, der simcha fun der man's farlangen." -bachophile | |
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| Axtremus | Mar 8 2006, 07:15 PM Post #14 |
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HOLY CARP!!!
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Area of any triangle with a right angle in it is always (height * base) / 2. I.e., area of triangle-with-right-angle is 1/2 of the enclosing rectangle. Therefore, area of rectangle is simply 6*8=48. No need to use Pythagoras theorem at all. (What's the proper name for "triangle with a right angle"? I cannot remember!) |
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| dolmansaxlil | Mar 8 2006, 07:26 PM Post #15 |
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HOLY CARP!!!
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ding ding ding! We have a winner!! |
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"Your first 10,000 photographs are your worst." ~ Henri Cartier-Bresson My Flickr Photostream | |
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| Horace | Mar 8 2006, 07:57 PM Post #16 |
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HOLY CARP!!!
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maybe an elegant way of explaining it would be to ask them what would happen if the paper was folded along the lines of the triangle... after they realize the paper would cover itself perfectly, it's a short logical step to realizing the area is twice that of the triangle. Edit - on second thought, it wouldn't cover itself. |
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| The 89th Key | Mar 8 2006, 08:52 PM Post #17 |
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Blast! Late to the party, but it's ok, I wasn't a math major. First off Dol, the kid should IMMEDIATELY realize that the length of the rectangle is 10. Period, end of story. If they dont know that within 30 seconds, they shouldn't be taking the test (using pyth's theorem). So of course, all you need left is the height of the rectangle. But like a few have said in here, you dont need to go that far. Once you have the area of the triangle (again, if they cant find that out, they shouldn't be taking the test - basic area of right trangle formula and all). Once they find out that the area is 24, they can see that with a vertical line, that the two triangles are just mirror images of each other, so just double the area of the main triangle.....the answer is 48. Ooops, and now I see that it's already been said multiple times. I'm a n00b. |
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| gryphon | Mar 8 2006, 09:18 PM Post #18 |
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Middle Aged Carp
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Because x = 90* you know that the triangle takes half of the rectangle. 6 x 8 = 48 (divide that by 2 but then multiply that by 2 for the total if you must) and you get 48. Again. There are a bunch of ways to solve this, but a good teacher uses problems that are easy to solve if you understand the concepts. This problem takes five seconds to solve if you understand it. I remember a calculus test problem that was very easy to solve if you fully understood the concept, but impossibly hard if you didn't. It simply involved subtracting two shapes from each other instead of trying to figure out complicated math equations to calculate the region. Kind of like dropping an irregular object in water to measure the displacement. |
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| maple | Mar 11 2006, 05:00 PM Post #19 |
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Junior Carp
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a minor point: the area of the triangle cxb is half the area of the rectangle abcd , regardless of the angle x. Likely, that is how the students derived the formula for the area of a triangle in the first place. To solve the problem they need to combine 2 facts: a) area of triangle = 1/2 * ax * bx b) area of triangle = 1/2 * area of rectangle if b) is not a "fact", they have to prove it by constructing the perpendicular xy on bc, and observing that the triangles xab and byx are congruent., as are xdc and cyx Try this with them, for a slightly longer deduction chain: In an isosceles right angled triangle ABC, with angle A being the right angle, let M and N be the middle of AB and AC respectively, and O the intersection of CM and BN. Determine the area of the quadrilateral AMON |
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