Respuesta :
1.
a.
A secant line is a line which intersects the circle at 2 different points.
A tangent line is a line which has only one point in common with the circle.
Check picture 1: The orange line s is a secant line, the blue line t is a tangent line.
b.
An inscribed angle is an angle formed by using 3 points of a circle.
The main property of an inscribed angle is that its measure is half of the measure of the arc it intercepts.
Check picture 2: If [tex]m(\angle KML)=\beta[/tex], then the measure of arc KL is [tex]2 \beta [/tex].
A central angle is an angle whose vertex is the center of the circle, and the 2 endpoints of the rays are points of the circle.
The main property is: the measure of the central angle is equal to the measure of the arc it intercepts.
Check picture 2
2.
To construct the inscribed circle of a triangle, we first draw the 3 interior angle bisectors of the triangle.
They meet at a common point called the incenter, which is the center of the inscribed circle.
We open the compass, from the incenter, so that it touches one of the sides at only one point. We then draw the circle. (picture 3)
To draw the circumscribed circle, we first find the midpoints of each side. We then draw perpendicular segments through these (the midpoints.) They meet at one common point, which is the circumcenter: the center of the circumscribed circle.
We open the compass from the circumcenter to one of the vertices of the triangle. We draw the circle, and see that it circumscribes the triangle.
(picture 4)
3.
Given an equation of a circle: [tex]x^2-2x+y^2+6y+6=0[/tex].
To determine the center and the radius of the equation we must write the above equation in the form :
[tex](x-a)^2+(y-b)^2=r^2[/tex].
Then, (a, b) is the center, and r is the radius of this circle. We do this process by completing the square.
Note that [tex]x^2-2x[/tex] becomes a perfect square by adding 1, and
[tex]y^2+6y[/tex] becomes a perfect square by adding 9.
Thus we have:
[tex]x^2-2x+y^2+6y+6=0\\\\(x^2-2x+1)+(y^2+6y+9)-4=0\\\\(x-1)^2+(y+3)^2=2^2[/tex]
Thus, the center is (1, -3), and the radius is 2.
4. Not complete
5.
The radius of the pizza is [tex]\displaystyle{ \frac{131}{2}ft=65.5ft [/tex].
The surface of a circle with radius r is given by the formula [tex]\displaystyle{ A=\pi r^2[/tex],
and the circumference is given by the formula C=2πr.
Thus, the area of the whole pizza is given by [tex]\displaystyle{ A=\pi r^2= \pi\cdot65.5^2=4290.25 \pi [/tex] (square ft).
Each of the 50 slices, has an area of [tex]\displaystyle{ \frac{4290.25 \pi}{50} =85.805 \pi [/tex] (square ft)
Notice that the perimeter (the crust) of a slice is made of 2 radii, and the arc-like part.
The arc is 1/50 of the circumference, so it is [tex] \displaystyle{\frac{2 \pi r}{50} = \frac{2\cdot65.5\cdot \pi }{50}= 2.62 \pi [/tex].
So the perimeter of one slice is 65.5+65.5+2.62π=131+2.62π
a.
A secant line is a line which intersects the circle at 2 different points.
A tangent line is a line which has only one point in common with the circle.
Check picture 1: The orange line s is a secant line, the blue line t is a tangent line.
b.
An inscribed angle is an angle formed by using 3 points of a circle.
The main property of an inscribed angle is that its measure is half of the measure of the arc it intercepts.
Check picture 2: If [tex]m(\angle KML)=\beta[/tex], then the measure of arc KL is [tex]2 \beta [/tex].
A central angle is an angle whose vertex is the center of the circle, and the 2 endpoints of the rays are points of the circle.
The main property is: the measure of the central angle is equal to the measure of the arc it intercepts.
Check picture 2
2.
To construct the inscribed circle of a triangle, we first draw the 3 interior angle bisectors of the triangle.
They meet at a common point called the incenter, which is the center of the inscribed circle.
We open the compass, from the incenter, so that it touches one of the sides at only one point. We then draw the circle. (picture 3)
To draw the circumscribed circle, we first find the midpoints of each side. We then draw perpendicular segments through these (the midpoints.) They meet at one common point, which is the circumcenter: the center of the circumscribed circle.
We open the compass from the circumcenter to one of the vertices of the triangle. We draw the circle, and see that it circumscribes the triangle.
(picture 4)
3.
Given an equation of a circle: [tex]x^2-2x+y^2+6y+6=0[/tex].
To determine the center and the radius of the equation we must write the above equation in the form :
[tex](x-a)^2+(y-b)^2=r^2[/tex].
Then, (a, b) is the center, and r is the radius of this circle. We do this process by completing the square.
Note that [tex]x^2-2x[/tex] becomes a perfect square by adding 1, and
[tex]y^2+6y[/tex] becomes a perfect square by adding 9.
Thus we have:
[tex]x^2-2x+y^2+6y+6=0\\\\(x^2-2x+1)+(y^2+6y+9)-4=0\\\\(x-1)^2+(y+3)^2=2^2[/tex]
Thus, the center is (1, -3), and the radius is 2.
4. Not complete
5.
The radius of the pizza is [tex]\displaystyle{ \frac{131}{2}ft=65.5ft [/tex].
The surface of a circle with radius r is given by the formula [tex]\displaystyle{ A=\pi r^2[/tex],
and the circumference is given by the formula C=2πr.
Thus, the area of the whole pizza is given by [tex]\displaystyle{ A=\pi r^2= \pi\cdot65.5^2=4290.25 \pi [/tex] (square ft).
Each of the 50 slices, has an area of [tex]\displaystyle{ \frac{4290.25 \pi}{50} =85.805 \pi [/tex] (square ft)
Notice that the perimeter (the crust) of a slice is made of 2 radii, and the arc-like part.
The arc is 1/50 of the circumference, so it is [tex] \displaystyle{\frac{2 \pi r}{50} = \frac{2\cdot65.5\cdot \pi }{50}= 2.62 \pi [/tex].
So the perimeter of one slice is 65.5+65.5+2.62π=131+2.62π
