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If you rotate that rectangle about the y-axis, you get the cylinder in the figure.
so i would need three other points?
Here is the given point, (0, 3)
okay so i would i have to use (0,0), (2,3), and (2,0)?
i believe that would make a rectangle and when its rotated, itll become a cylinder
how did you determine the rest of the dimensions when you were only given one point?
Excellent job! Are you sure you need to ask me questions. Maybe I should be asking you to help me.
When a rectangle that has an edge on the y-axis rotates about the y-axis, it creates a cylinder.
if you plug that all into geogebra or another graphing thing, you'll be able to see that it forms a rectangle
was i correct to do that??
Yes. Good job!
The problem asked for points, not equations. Also your triangle will yield a cone of rotation, not a cylinder.
good job at pointing out the concept, regardless
Quadrilateral OPQR is inscribed inside a circle as shown below. Write a proof showing that angles O and Q are supplementary. Here is the question again
i believe that i just needed to provide the points.
cylinder is simply a cone, particularly when we are talking about projective geometry
Look at the figure above. It shows circle O and a central angle of 80 degrees. What is x, the arc measure of the arc intercepted by the central angle of 80 degrees?
im not sure
Ok, no problem. The arc measure is the same as the central angle measure. Since the central angle is 80 deg, then x is also 80 deg.
Ok so far?
ohhh okay yes, i understand
That was the case of a central angle. Now we need the case of an inscribed angle. An inscribed angle is an angle whose vertex is on the circle itself.
In the case of an inscribed angle, the measure of the intercepted arc is twice the measure of the inscribed angle.
Are you ok with central angle and arc & inscribed angle and arc?
i believe so
im just a bit slow rn
its 12:40AM where i am rn
Here's a summary: central angle measure = arc measure inscribed angle measure = 1/2 arc measure
Ok. I'll just finish this problem and that 'll be it for tonight.
and this is for the Quadrilateral OPQR is inscribed inside a circle question?
Yes. Now let's get back to our inscribed quadrilateral.
lets do it
Here is the inscribed quadrilateral. Let's call the circle circle C with center at point C.
If you use CA as the side of a central angle, and you go around a full rotation until you end up again at CA, that would be a 360-deg angle, right?
Then since that central angle measures 360 degrees, that means that a circle is considered to be an arc of measure 360 deg. Remember, we saw before that the arc measure is the same as the central angle measure.
Now let's deal with angles O and Q to prove they are supplementary.
so hold on just a second
how should i write this for the answer?
I'm doing it now.
but in words
The two arcs, x and y, add up to the full circle, so their measures add up to 360 deg. x + y = 360
If we solve for y, we get y = 360 - x That means we have two arcs of measures x and 360 - x.
|dw:1439787195206:dw| Now let's look at angle Q. Angle Q is an inscribed angle, An inscribed angle is half of its intercepted arc. The intercepted arc of angle Q is arc ROP Since arc ROP has measure x, then the inscribed angle Q has half of that measure, 1/2 x
|dw:1439787388301:dw| Now we do the same with angle Q. Angle Q is an inscribed angle. Angle Q intercepts arc PQR. The measure of angle Q is half the measure of angle PQR. That means the measure of angle Q is 1/2 (360 - x) = 180 - 1/2 x
Now that we have measures for angles O and Q, we recall the definition of supplementary angles: Two angles are supplementary if the sum of their measures is 180 deg.
Since we want to show that angle O and angle Q are supplementary angles, we add their measures. m
Ok, gtg, it's very late for me. If you have any questions, just ask. I'll try to answer them tomorrow.