Difference between revisions of "2006 AMC 8 Problems/Problem 5"
Basketball8 (talk | contribs) (→Solution) |
(→Solution) |
||
| Line 18: | Line 18: | ||
== Solution == | == Solution == | ||
| + | ===Solution 1=== | ||
| + | Drawing segments <math>AC</math> and <math>BD</math>, the number of triangles outside square <math>ABCD</math> is the same as the number of triangles inside the square. Thus areas must be equal so the area of <math>ABCD</math> is half the area of the larger square which is <math> \frac{60}{2}=\boxed{\textbf{(D)}\ 30 } </math>. | ||
| + | |||
| + | ===Solution 2=== | ||
If the side length of the larger square is <math> x </math>, the side length of the smaller square is <math> \frac{\sqrt{2} \cdot x}{2} </math>. Therefore the area of the smaller square is <math> \frac{x^2}{2} </math>, half of the larger square's area, <math> x^2 </math>. | If the side length of the larger square is <math> x </math>, the side length of the smaller square is <math> \frac{\sqrt{2} \cdot x}{2} </math>. Therefore the area of the smaller square is <math> \frac{x^2}{2} </math>, half of the larger square's area, <math> x^2 </math>. | ||
Thus, the area of the smaller square in the picture is <math> \frac{60}{2}=\boxed{\textbf{(D)}\ 30 } </math>. | Thus, the area of the smaller square in the picture is <math> \frac{60}{2}=\boxed{\textbf{(D)}\ 30 } </math>. | ||
| + | |||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2006|num-b=4|num-a=6}} | {{AMC8 box|year=2006|num-b=4|num-a=6}} | ||
Revision as of 18:53, 24 December 2012
Problem
Points 
and 
are midpoints of the sides of the larger square. If the larger square has area 60, what is the area of the smaller square?
![[asy]size(100); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle,linewidth(1)); draw((0,1)--(1,2)--(2,1)--(1,0)--cycle); label("$A$", (1,2), N); label("$B$", (2,1), E); label("$C$", (1,0), S); label("$D$", (0,1), W);[/asy]](http://latex.artofproblemsolving.com/e/7/8/e7899c538a174faf9963a60d09886b7fb12ae942.png)
Solution
Solution 1
Drawing segments 
and 
, the number of triangles outside square 
is the same as the number of triangles inside the square. Thus areas must be equal so the area of is half the area of the larger square which is 
.
Solution 2
If the side length of the larger square is 
, the side length of the smaller square is 
. Therefore the area of the smaller square is 
, half of the larger square's area, 
.
Thus, the area of the smaller square in the picture is .
See Also
| 2006 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 | ||
