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 product of its diagonals, and the diagonals are the dimensions of the rectangle, so the area of the kite is <math> \frac{(18)(21)}{2}=189 </math>. Thus, the area of the remaining gold is <math> 378-189=189, \boxed{\text{D}} </math>.
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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 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.

[asy] for (int a = 0; a < 7; ++a) { for (int b = 0; b < 8; ++b) { dot((a,b)); } }  draw((3,0)--(0,5)--(3,7)--(6,5)--cycle); [/asy]

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?

$\text{(A)}\ 63 \qquad \text{(B)}\ 72 \qquad \text{(C)}\ 180 \qquad \text{(D)}\ 189 \qquad \text{(E)}\ 264$

Solution

The large grid has dimensions three times that of the small grid, so its dimensions are $3(6)\times3(7)$, or $18\times21$, so the area is $(18)(21)=378$. The area of the kite is half of the area of the rectangle as you can see. Thus, the area of the kite is $378/2=189, \boxed{\text{D}}$.

See Also

2001 AMC 8 (ProblemsAnswer KeyResources)
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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