Difference between revisions of "2008 AMC 10B Problems/Problem 10"
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==Solution== | ==Solution== | ||
− | + | Let the center of the circle be O. Draw lines | |
− | + | OA, OB, and OC. | |
− | + | OA=OB=5, since they are both radii. | |
− | + | OC bisects AB, and AB=6, so letting point D be | |
− | + | the intersection of AB and OC, AD=BD=3. | |
− | + | Also, OD=4. This means that CD=1. | |
− | + | Using the pythagorean theorem, | |
+ | |||
+ | <math>AC=\sqrt{3^2+1^2}=\sqrt{10}</math>. | ||
+ | |||
+ | Answer A is the correct answer. | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2008|ab=B|num-b=9|num-a=11}} | {{AMC10 box|year=2008|ab=B|num-b=9|num-a=11}} |
Revision as of 17:41, 10 August 2008
Problem
Points A and B are on a circle of radius 5 and AB=6. Point C is the midpoint of the minor arc AB. What is the length of the line segment AC?
Solution
Let the center of the circle be O. Draw lines
OA, OB, and OC.
OA=OB=5, since they are both radii.
OC bisects AB, and AB=6, so letting point D be
the intersection of AB and OC, AD=BD=3.
Also, OD=4. This means that CD=1.
Using the pythagorean theorem,
.
Answer A is the correct answer.
See also
2008 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
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 AMC 10 Problems and Solutions |