We're given the wedge of a circle with an angle, at the tapered end, of 60º. Inscribed in the wedge is a smaller circle whose area is given as π cm^2. What, then, is the area of the leftover space in the wedge?
The wedge is obviously 1/6 of a whole circle if the wedge's tapered angle is 60º (60º is 1/6 of 360º). To calculate the wedge's area, we need to know the radius of the circle that the wedge belongs to.
If the small circle has an area of π, then we know that the radius of the small circle is 1 cm. (A = πr^2; r = 1 cm) We can then draw perpendicular line segments to the three points of tangency with the wedge. These line segments all equal 1 cm.
Next, we can draw a segment connecting the center of the small circle to the wedge's tapered angle. This segment bisects the 60º angle, turning it into two 30º angles. In turn, we now have two 30-60-90 right triangles inside the figure forming a "kite" shape.
The ratio of side lengths for a 30-60-90 triangle are 1:√3:2 (leg:leg:hypotenuse). So we can see that the length of the segment connecting the vertex of the tapered angle to the small circle's center is 2 cm, and from the circle's center to the far side of the wedge is another 1 cm. So the diameter of the imaginary large circle containing the wedge is 3.
Now, it's just a matter of calculating the area of the wedge, then subtracting the area of the small circle, which is π, to get our answer. The area of the (imaginary) large circle is 9π cm^2, so the wedge's area is 9π/6, or 3π/2 cm^2. If we subtract the area of the smaller circle, π, from 3π/2, we get
3π/2 - π = π/2
So the answer is that the leftover area inside the wedge equals π/2.
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We're given the wedge of a circle with an angle, at the tapered end, of 60º. Inscribed in the wedge is a smaller circle whose area is given as π cm^2. What, then, is the area of the leftover space in the wedge?
ReplyDeleteThe wedge is obviously 1/6 of a whole circle if the wedge's tapered angle is 60º (60º is 1/6 of 360º). To calculate the wedge's area, we need to know the radius of the circle that the wedge belongs to.
If the small circle has an area of π, then we know that the radius of the small circle is 1 cm. (A = πr^2; r = 1 cm) We can then draw perpendicular line segments to the three points of tangency with the wedge. These line segments all equal 1 cm.
Next, we can draw a segment connecting the center of the small circle to the wedge's tapered angle. This segment bisects the 60º angle, turning it into two 30º angles. In turn, we now have two 30-60-90 right triangles inside the figure forming a "kite" shape.
The ratio of side lengths for a 30-60-90 triangle are 1:√3:2 (leg:leg:hypotenuse). So we can see that the length of the segment connecting the vertex of the tapered angle to the small circle's center is 2 cm, and from the circle's center to the far side of the wedge is another 1 cm. So the diameter of the imaginary large circle containing the wedge is 3.
Now, it's just a matter of calculating the area of the wedge, then subtracting the area of the small circle, which is π, to get our answer. The area of the (imaginary) large circle is 9π cm^2, so the wedge's area is 9π/6, or 3π/2 cm^2. If we subtract the area of the smaller circle, π, from 3π/2, we get
3π/2 - π = π/2
So the answer is that the leftover area inside the wedge equals π/2.
QED.