Difference between revisions of "2001 AMC 8 Problems/Problem 9"

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draw((3,0)--(0,5)--(3,7)--(6,5)--cycle);
 
draw((3,0)--(0,5)--(3,7)--(6,5)--cycle);
 
</asy>
 
</asy>
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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?
 
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?
  

Revision as of 03:21, 14 January 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 the product of its diagonals, and the diagonals are the dimensions of the rectangle, so the area of the kite is $\frac{(18)(21)}{2}=189$. Thus, the area of the remaining gold is $378-189=189, \boxed{\text{D}}$.

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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