Difference between revisions of "2006 UNCO Math Contest II Problems/Problem 1"
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==Solution== | ==Solution== | ||
| − | To solve this we start by breaking it up into it's four shaded areas. Starting with the bottom right one, it splits four squares | + | To solve this we start by breaking it up into it's four shaded areas. Starting with the bottom right one, we see it splits four squares. It hits the vertex on the left end, and it hits both vertexes on the right end so we know that the total area of that region in two squares. The reason for this is because we can take the triangles and rearrange them into two squares. |
==See Also== | ==See Also== | ||
Revision as of 16:01, 23 December 2014
Problem
If a dart is thrown at the 
target, what is the probability that it will hit the shaded area?
![[asy] filldraw((2,2)--(4,6)--(6,6)--(6,4)--cycle,blue); filldraw((2,2)--(6,2)--(6,1)--cycle,blue); filldraw((2,2)--(0,0)--(0,1)--cycle,blue); filldraw((2,2)--(0,4)--(0,6)--cycle,blue); for(int i=0;i<7;++i){ draw((0,i)--(6,i),black); draw((i,0)--(i,6),black); } dot((2,2));dot((0,4));dot((0,6));dot((4,6));dot((6,6)); dot((6,4));dot((6,2));dot((6,1));dot((0,0));dot((0,1)); [/asy]](http://latex.artofproblemsolving.com/d/f/1/df1c252b085c82d7d67511a0e58cc24d9595ae8a.png)
Solution
To solve this we start by breaking it up into it's four shaded areas. Starting with the bottom right one, we see it splits four squares. It hits the vertex on the left end, and it hits both vertexes on the right end so we know that the total area of that region in two squares. The reason for this is because we can take the triangles and rearrange them into two squares.
See Also
| 2006 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by First Question |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
