2000 AMC 8 Problems/Problem 6: Difference between revisions
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Square FECG¡ square FHIJ = <math>4 \times 4 - 3 \times 3 = 16 - 9 = \boxed{\text{(B) 7}}</math>. | Square FECG¡ square FHIJ = <math>4 \times 4 - 3 \times 3 = 16 - 9 = \boxed{\text{(B) 7}}</math>. | ||
==See Also== | |||
{{AMC8 box|year=2000|num-b=5|num-a=7}} | |||
Revision as of 09:17, 15 May 2011
Problem
Figure 
is a square. Inside this square three smaller squares are drawn with the side lengths as labeled. The area of the shaded 
-shaped region is
![[asy] pair A,B,C,D; A = (5,5); B = (5,0); C = (0,0); D = (0,5); fill((0,0)--(0,4)--(1,4)--(1,1)--(4,1)--(4,0)--cycle,gray); draw(A--B--C--D--cycle); draw((4,0)--(4,4)--(0,4)); draw((1,5)--(1,1)--(5,1)); label("$A$",A,NE); label("$B$",B,SE); label("$C$",C,SW); label("$D$",D,NW); label("$1$",(1,4.5),E); label("$1$",(0.5,5),N); label("$3$",(1,2.5),E); label("$3$",(2.5,1),N); label("$1$",(4,0.5),E); label("$1$",(4.5,1),N); [/asy]](http://latex.artofproblemsolving.com/7/c/6/7c6ea431a00e38c9ba0b8513fa0b9a39252cfcef.png)
Solution
Square FECG¡ square FHIJ = 
.
See Also
| 2000 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 | ||
