Difference between revisions of "2003 AMC 8 Problems/Problem 25"
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==Solution== | ==Solution== | ||
− | The side lengths of square <math>\text{WXYZ}</math> must be 5 cm, since the area is <math>25 {cm}^2</math>. First, you should determine the height of <math>\triangle{ABC}</math>. The distance from <math>\text{O}</math> to line <math>\text{WZ}</math> must be 2.5 cm, since line <math>\text{WX}</math> = 5 cm, and the distance from <math>\text{O}</math> to <math>\text{Z}</math> is half of that. The distance from line <math>\text{WZ}</math> to line <math>\text{BC}</math> must be 2, since the side lengths of the small squares are 1, and there are two squares from line <math>\text{WZ}</math> to line <math>\text{BC}</math>. So, the height of <math>\triangle{ABC}</math> must be 4.5, which is 2.5 + 2. The length of <math>\text{BC}</math> can be determined by subtracting 2 from 5, since the length of <math>\text{WZ}</math> is 5, and the two squares in the corners give us 2 together. This gives us the base for <math>\triangle{ABC}</math>, which is 3. Then, we multiply 4.5 by 3 and divide by 2, to get an answer of <math>{(C)}\ \frac{27}{4}</math>. | + | The side lengths of square <math>\text{WXYZ}</math> must be 5 cm, since the area is <math>25\ {cm}^2</math>. First, you should determine the height of <math>\triangle{ABC}</math>. The distance from <math>\text{O}</math> to line <math>\text{WZ}</math> must be 2.5 cm, since line <math>\text{WX}</math> = 5 cm, and the distance from <math>\text{O}</math> to <math>\text{Z}</math> is half of that. The distance from line <math>\text{WZ}</math> to line <math>\text{BC}</math> must be 2, since the side lengths of the small squares are 1, and there are two squares from line <math>\text{WZ}</math> to line <math>\text{BC}</math>. So, the height of <math>\triangle{ABC}</math> must be 4.5, which is 2.5 + 2. The length of <math>\text{BC}</math> can be determined by subtracting 2 from 5, since the length of <math>\text{WZ}</math> is 5, and the two squares in the corners give us 2 together. This gives us the base for <math>\triangle{ABC}</math>, which is 3. Then, we multiply 4.5 by 3 and divide by 2, to get an answer of <math>\boxed{\textbf{(C)}\ \frac{27}{4}}</math>. |
+ | ==See Also== | ||
{{AMC8 box|year=2003|num-b=24|after=Last Problem}} | {{AMC8 box|year=2003|num-b=24|after=Last Problem}} |
Revision as of 03:22, 24 December 2012
Problem
In the figure, the area of square is . The four smaller squares have sides 1 cm long, either parallel to or coinciding with the sides of the large square. In , , and when is folded over side , point coincides with , the center of square . What is the area of , in square centimeters?
Solution
The side lengths of square must be 5 cm, since the area is . First, you should determine the height of . The distance from to line must be 2.5 cm, since line = 5 cm, and the distance from to is half of that. The distance from line to line must be 2, since the side lengths of the small squares are 1, and there are two squares from line to line . So, the height of must be 4.5, which is 2.5 + 2. The length of can be determined by subtracting 2 from 5, since the length of is 5, and the two squares in the corners give us 2 together. This gives us the base for , which is 3. Then, we multiply 4.5 by 3 and divide by 2, to get an answer of .
See Also
2003 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |