Triangle ABO is an isosceles right triangle, AC is a
diameter of circle O, and AB is a diameter of circle
D. If the semicircle centered at O has area 2π, what
is the shaded area AEBF (which is called a lune)?
To solve this problem, we can start by finding the area of the larger triangle ABO and the area of the semicircle centered at O.
Triangle ABO is an isosceles right triangle, so we know that angle ABO is 45 degrees. Since angle ABO is the angle inscribed in circle D, it is also an inscribed angle of a semicircle. By the Inscribed Angle Theorem, we know that angle ABO is half the measure of the central angle.
Since the central angle in a semicircle is 180 degrees, angle ABO is 180/2 = 90 degrees. Therefore, triangle ABO is a 45-45-90 triangle.
In a 45-45-90 triangle, the ratio of the lengths of the sides is 1:1:√2. Since AB is a diameter of circle D, it is equal to twice the length of AO. Therefore, the length of AO is AB/2.
Using the ratio of the sides in a 45-45-90 triangle, we know that AO/AB = 1/√2. Rearranging this equation, we have AO = AB/√2.
Since AC is a diameter of circle O, it is equal to twice the length of CO. Therefore, CO = AC/2.
Let's assume the radius of circle D is r. Then, the radius of circle O is r/√2.
The area of a semicircle is equal to half the area of a full circle, so the area of the semicircle centered at O is (1/2)(π(r/√2)^2) = πr^2/4.
We are given that the area of the semicircle is 2π, so we can set up the equation:
πr^2/4 = 2π
Multiplying both sides by 4, we have:
πr^2 = 8π
Dividing both sides by π, we get:
r^2 = 8
Taking the square root of both sides, we have:
r = √8 = 2√2
Now, we can find the length of AO:
AO = AB/√2 = (2√2)/√2 = 2
Therefore, the area of triangle ABO is (1/2)(AB)(AO) = (1/2)(2)(2) = 2 square units.
Now, let's find the shaded area AEBF, which is called a lune.
The lune is formed by the arc AB of circle D and the arc EF of circle O.
The length of arc AB is half the circumference of circle D, which is (1/2)(2πr) = 2π.
The length of arc EF is half the circumference of circle O, which is (1/2)(2π(r/√2)) = (1/2)(2π(2√2/√2)) = 2π(2) = 4π.
The lune is the area between these two arcs. To find the shaded area, we can subtract the area bounded by the arc EF from the area bounded by the arc AB.
Therefore, the shaded area AEBF is 2π - 4π = -2π.
Note that the shaded area is negative because the arc EF is larger than the arc AB.
Therefore, the shaded area AEBF is -2π square units.