Divided Rectangle - Sept. 2-6, 1996
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A rectangle is divided into four rectangles with areas 45, 25, 15, and x. Find x.
- Annie Fetter
Annie says: Solutions
Here we have it, the first problem of the new school year in the U.S. We got 84 correct solutions to this problem and only 5 incorrect. That's pretty good!
My first comment applies to a number of people: include an explanation! You must explain how you got the answer. Pretend that you are writing a solution that will be read by someone who doesn't understand how to do the problem. That way you'll be sure that you understand the problem completely, and you'll have an easier time when it comes to solving harder problems.
The "proportional" method is a really good one - a number of people used this, so look out for it. Don't assume that one figure is a square and hence the factors of 25 therefore have to be 5 and 5. It turns out that 25 and 1 don't work, but be careful what you assume from the picture.
Jenifer Rajkumar pointed out all the assumptions she made in solving the problem, which was great. I'm sure a lot of you made the same assumptions without even thinking. Michael Thwaites made a very good point and you should read his submission - it's not an answer, but it could well be a POW some day.
Jason Lee and Kristin Hartono showed a really nice way to check their work, always a good idea. Ken Duisenberg made an excellent point about the picture vs. the text - looking at the picture in conjunction with the text means the answer will have to be one thing, but if you simply read the text and don't have the picture, things get a little crazy.
Below are some exemplary solutions. A list of all the folks who got this problem right and most of the solutions are also available.
Jason Lee Grade: 9 School: Stuyvesant High School, Manhattan, New York ________b______________a____ | | | | | | c 45 c 25 c | | | |________b_________|____a____| | | | d X d 15 d |________b_________|____a____| A rectangle is divided into four rectangles with areas 45, 25, 15, and x. Find x. I have labeled the line segments a, b, c and d. The LCD of 25 and 15 is 5. (Even though 25 and 1 work too) a would thus be 5. That also makes c equal to 5 (25/5) and d equal to 3 (15/5). Then b would have to be 9 (45/5). a=5, b=9, c=5, d=3 b * d = x 9 * 3 = x 27 = x Then checking myself: 45 + 25 + 15 + 27 = 112 a + b = L c + d = W A = L * W 5 + 9 = L 5 + 3 = W A = 14 * 8 14 = L 8 = W A = 112 Kristin Yuliana Hartono
Grade: 8 School: Raffles Girls' Secondary School, Singapore Length of the sides of rectangle with area of 25 is 5. 15/3 = 5, so the width of the rectangle with the area of 15 is 5. 45/5= 9, so the length of the rectangle with the area of 45 is 5 and its width is 9. Thus, the length of rectangle with area of x is 9 and its width is 3. x = 9 x 3 = 27 To check whether the answer is correct: Length of the big rectangle is 14 and its width is 8, thus its area is 112 (14 x 8). Total areas of the small rectangles (including x) = 45 + 25 + 15 + 27 = 112 ANSWER: x = 27
Tomoni Nakajima Grade: ? - Intensive Geometry Period 2 School: Newport High School, Bellevue, Washington The area of the rectangle is 27. Step 1: Find the Lowest Common Multiple of 25, 15, and 45, which is 5. Step 2: Since the area of a rectangle and square can be found by using this formula: (length) x (width) = area let x be either length of width depending on the square or rectangle. Step 2(a): The width of the rectangle that has an area of 45 is 5. Calculate: 45 = 5(x) x = 9 Therefore, the length of rectangle that has an area of 45 is 9. Step 2(b): The width of the square that has an area of 25 is 5. Calculate: 25 = 5(x) x = 5 Therefore, the length of the square is 5. Step 2(c): The length of the rectangle that has an area of 15 is 5. Calculate: 15 = 5(x) x = 3 Therefore, the width of this rectangle is 3. Step 3: From step 2(a)(b)(c), now I know that the length of the rectangle that has the area of (x) is 9, and the width is 3. area = (length) x (width) = (9) x (3) = 27
Thomas Kuo Grade: 9 School: Burroughs High School, Ridgecrest, Calfornia A E B * * * * * * * * * * * * * * * * * * * * * * * * * * 45 * 25 * * * * H * * * * * * * * * * * * * * * * * * * * * * F * I * * * * * * x * 15 * * * * * * * * * * * * * * * * * * * * * * * * * * * * * D G C ABCD is a rectangle. The area of x has an area of 27. (1) Find AH, EI, and BF. A common factor of 45 and 25 is 5. So AH, EI, and BF could equal 5. (2) Find EB, IF, and GC. The area of EBFI, 25, divded by the BF, 5, is 5, making EB, IF, and GC equal to 5. (3) Find AE, HI, and DG. The area of AEIH, 45, divided by the length of AH, 5, is equal to 9, so AE, HI, and DG are equal to 9. (4) Find x. Since HI = 9 and HD = 3, the area of x = 27 and since there can be only one area for that region, x can only equal 27.
Jenifer Rajkumar Grade: ? School: ? Making the following three assumptions, the area of X is 27: 1. The problem is dealing with integers. 2. The rectangle with the area of 25 is a square. 3. The rectangle with the area of 15 is wider than it is high. However, if you do not assume anything, the answers to the problem are infinite.
Ken Duisenberg Grade: Hewlett Packard Engineer School: Stanford University '93 Answer: 27 (or 75) Solution: If we take the diagram as the statement of the problem, then we simply have to find the dimensions of the smaller rectangles to find the dimensions (and thus the area) of X. 25 is either 1x25 or 5x5, but since 25 is not a factor of either 45 or 15, it must be 5x5. So the remaining dimensions of X must be 3 (shared with 15) and 9 (shared with 45). So the area of X is 27. If, however, we just use the text as the statement of the problem, there is a second way the rectangle could be drawn, with the area of 15 opposite X. Then 15 shares a side of 5 with 25 and a side of 3 with 45. The dimensions of X are then 15 (shared with 45) and 5 (shared with 25). So the area of X is 75. [There is also the possibility that the four rectangles do not meet at the same internal point, in which case there are an infinite number of areas for X.]
Michael Thwaites School: lovely Northern California Annie - This is interesting because knowing the value of x doesn't tell me the sides of the rectangles. However, you get a unique solution if the sides of the four smaller rectangles (and thus the large one too) are integers. What if we only require the big rectangle to have integer sides and the smaller to have, say, rational sides? Can we use this idea to make a bigger better puzzle? Michael
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