A,B,C,D,E,F are 6 consecutive points on the circumference of a circle such that AB=BC=CD=10,DE=EF=FA=22. If the radius of the circle is √n, what is the value of n?
Details and assumptions:
-The lengths given are side lengths, not arcs.
Let the center of the circle be O.
Let 2θ = ∠AOB
Let 2φ = ∠EOF
6θ+6φ = 2π
θ+φ = π/3
sin(θ+φ) = √3/2
Now,
in ∆AOB, sinθ = 5/r
in ∆EOF, sinφ = 11/r
sin(θ+φ) = sinθcosφ + cosθsinφ
so
5/r * √(r^2-121)/r + 11/r * √(r^2-25)/r = √3/2
5√(r^2-121) + 11√(r^2-25) = √3/2 r^2
r^2 = 268
To find the value of n, we need to use some geometric properties of a circle.
Let's start by drawing the circle and labeling the points A, B, C, D, E, and F on the circumference.
F E
A D
B C
We are given that AB = BC = CD = 10 and DE = EF = FA = 22.
Now, we know that the lengths of the chords AB, BC, CD, DE, EF, and FA are equal, which makes them congruent chords. In a circle, congruent chords are equidistant from the center.
Let's name the center of the circle O. Since AB, BC, and CD are congruent chords, they must be equidistant from the center. Similarly, DE, EF, and FA are also equidistant from O.
Let's consider the chord AB. Since AB is equidistant from the center O, the perpendicular bisector of AB passes through O. The same applies to CD and the perpendicular bisector of CD.
Now, let's connect the midpoints of AB, BC, CD, DE, EF, and FA with straight lines. These lines bisect the chords and are perpendicular to them. The shape formed by connecting these points is a regular hexagon.
A-----------M-----------D
/ \
/ \
/ \
B C
| |
| |
---O------------------E
In the regular hexagon, the distance from the center O to any vertex (such as A, B, C, D, E, or F) is equal to the radius of the circle. Let's consider the triangle OAM.
Triangle OAM is an equilateral triangle since all sides of the hexagon are congruent. Therefore, the measure of each angle in OAM is 60 degrees.
We can then use the Law of Cosines to find the length of OA or the radius of the circle. The Law of Cosines states:
c^2 = a^2 + b^2 - 2ab * cos(C)
For triangle OAM, we have c = OA, a = OM, and b = AM. Since it's an equilateral triangle, a = b.
OA^2 = OM^2 + OM^2 - 2(OM)(OM) * cos(60°)
OA^2 = 2OM^2 - 2(OM)^2 * cos(60°)
OA^2 = OM^2
Since OM is half of AB, which is 10, OM = 5.
OA^2 = 2(5)^2 - 2(5)^2 * cos(60°)
OA^2 = 50 - 50 * 0.5
OA^2 = 50 - 25
OA^2 = 25
Therefore, the radius of the circle is √25 = 5.
Since the question asks for the value of n, which is the square of the radius, n = 5^2 = 25.
So, the value of n is 25.