Difference between revisions of "2007 AMC 10B Problems/Problem 11"
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Let the radius of the circle = <math>r</math>. | Let the radius of the circle = <math>r</math>. | ||
Let the distance between origin and base of triangle = <math>a</math>. | Let the distance between origin and base of triangle = <math>a</math>. | ||
− | 1 + a^2 = r^2 r + a = 2sqrt(2) a = (2)sqrt(2) - r 9 - (4r)sqrt(2) = 0 r = ((9)sqrt(2))/8 | + | 1 + a^2 = r^2 r + a = 2sqrt(2) a = (2)sqrt(2) - r 9 - (4r)sqrt(2) = 0 r = ((9)sqrt(2))/8 \pi r^2 = 81\pi/32 |
== See also == | == See also == |
Revision as of 15:17, 11 July 2020
Contents
Problem
A circle passes through the three vertices of an isosceles triangle that has two sides of length and a base of length . What is the area of this circle?
Solutions
Solution 1
Let have vertex and center , with foot of altitude from at .
Then by Pythagorean Theorem (with radius and height ) on
Substituting and solving gives . Then the area of the circle is .
Solution 2
By (or we could use and Heron's formula), and the answer is
Alternatively, by the Extended Law of Sines, Answer follows as above.
Solution 3
Extend segment to on Circle .
By the Pythagorean Theorem
is similar to , so which gives us therefore
The area of the circle is therefore
Solution 4
First, we extend to hit the circle at
By the Pythagorean Theorem, we know that We also know that, from the Power of a Point Theorem, We can substitute the ones we know to get We can simplify this to get that We add and together to get the length of the diameter, and then we can find the area. Therefore, the radius is , so the area is
Solution 5
Another possible solution is to plot the circle and triangle on a graph with the circle having center (0,0). Let the radius of the circle = . Let the distance between origin and base of triangle = .
1 + a^2 = r^2 r + a = 2sqrt(2) a = (2)sqrt(2) - r 9 - (4r)sqrt(2) = 0 r = ((9)sqrt(2))/8 \pi r^2 = 81\pi/32
See also
2007 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.