Difference between revisions of "2001 AMC 8 Problems/Problem 9"
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− | The large grid has dimensions three times that of the small grid, so its dimensions are <math> 3(6)\times3(7) </math>, or <math> 18\times21 </math>, so the area is <math> (18)(21)=378 </math>. The area of the kite is half | + | The large grid has dimensions three times that of the small grid, so its dimensions are <math> 3(6)\times3(7) </math>, or <math> 18\times21 </math>, so the area is <math> (18)(21)=378 </math>. The area of the kite is half of the area of the rectangle as you can see. Thus, the area of the kite is <math> 378/2=189, \boxed{\text{D}} </math>. |
==See Also== | ==See Also== | ||
{{AMC8 box|year=2001|num-b=8|num-a=10}} | {{AMC8 box|year=2001|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:35, 18 April 2021
Problem
To promote her school's annual Kite Olympics, Genevieve makes a small kite and a large kite for a bulletin board display. The kites look like the one in the diagram. For her small kite Genevieve draws the kite on a one-inch grid. For the large kite she triples both the height and width of the entire grid.
The large kite is covered with gold foil. The foil is cut from a rectangular piece that just covers the entire grid. How many square inches of waste material are cut off from the four corners?
Solution
The large grid has dimensions three times that of the small grid, so its dimensions are , or , so the area is . The area of the kite is half of the area of the rectangle as you can see. Thus, the area of the kite is .
See Also
2001 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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 |
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