how-do-you-see-it-compare-the-circumference-of-odot-mathrmp-to-the-length-of-de-explain-your-reasoni

Question

HOW DO YOU SEE IT?
Compare the circumference of $\odot \mathrm{P}$ to the length of DE. Explain your reasoning.

Georgiann Andersen · Accepted Answer

Okay, we&#x27;re gonna do problem. 36. They have a circle with a diameter of d E. And this is circle C inside. That they have a smaller circle is a circle p. They want to know the comparison of the circumference of a circle p to the length of our d. So on circle P, let&#x27;s just make the radius. Are so our circumference a circle p we&#x27;ll be two pi r. Well, our radius of circle C then we&#x27;ll be to our. So that means that our circumference of circle C will be two times pi times to our which would be the same thing as saying for our pie. Okay, so we know that the measure then of D E is 100 and 80 degrees because that&#x27;s half of the circle. We know that he&#x27;s the diameter. So to find the ark link, we&#x27;ll take the 180 degrees and divide that by 3 60 multiply by the circumference which we found was for pi r. And this will turn out to be 1/2 times four pi r, which is gonna end up being two pi r and noticed that the circumference of point key equals two pi R So the arc length is the same arc length. Ah, Arc de e equals the circumference of point P Oh!

Raushan Kumar · Answer

We&#x27;re going to do problem with 67 in this worsen. Or you could do we have to compare the length of radius and diameter. So first of all, little savory really is. And what is damaged releases the distance from center to any point on the circle. So this is our nobody. Demeter. Demeter is the lent off two points. Where did you put it To each other and the lane person and touching each others would pass through the center. So this becomes the diameter. Now we know what is the radios and what is diameter. So we can say that radius will be equal to half of diameter. Okay, as boot passes to the center radius and diameter boat Ah, pass through center. So we can say that releases equals Toby by toe, as from Tyga, we can notice

Palitha Reddy · Answer

So for this question, were pretty much asked to compare the different sizes of the two angles that are given to us. So in general, when we&#x27;re trying to compare, you know, angles, we can Onley compare the measures of the angles the degrees. So just because, you know, the rays are larger in one angle and smaller and the other angle, it doesn&#x27;t mean that I&#x27;m gonna take that into account, because whenever we&#x27;re doing angle measures right, we could only take that number that angle value into consideration. So in this case, right angle P and angle. Q. So both of these arcs, right? Thes decree measures are going to be the exact sake, even if the&#x27;s raise, even though the rays of PR definitely much smaller right in the rays of Q. We&#x27;re not gonna take that into account when considering through the angle measures, and basically, if these two are equal or not, so in this case, the angle size or the angle measure of pink, you are going to be equal despite the size off their raise. You know, like another example is when we&#x27;re thinking about a right angle, it doesn&#x27;t matter if it&#x27;s this big right against this big. Or if it&#x27;s this big, they&#x27;re both going to be 90 degrees so similar.

Natalie Daly · Answer

Okay, so for this problem were given that, uh, the area of a circle is equal to 36 pi into square, and we&#x27;re told to find the east your comfort to the circle. Our first step to doing this will be to find the radius of the circle, because we can use that to find the circumference and in the chapter were given this hopeful creation of area is equal to high our twit. Um so what we can use plug in this value of, Ah, the area is equal to 36. Pie equals pi r squared. Uh, and then we consult for our so we can eliminate the pie on both sides, get 36 equal to R squared and are equal to the square root of 36 which is equivalent to six inches to a radius of six inches. Ah. Then you can use this other helpful equation from the chapter. Your conference is equal to two high times the radius. So we know that the ship conference will be equal to two pi from six, which is equal to 12 high. So this your conference of the circle is Terry. It&#x27;s about pie. So the circumference of the circle, um, will be equal to 12 pie inches, and that&#x27;s how you solved for the issue conference.

Rishi Kavikondala · Answer

this problem. We have to solve the formula given in the previous question. So the circum circumference formula for pi so dividing both sides by two are since we&#x27;re gonna solve for pie, you get high is equal to see over to our we&#x27;re told in the problem. We have to notice why pious sometimes defined as the ratio of the circumference to the diameter. So circumference is obviously the numerator, so that&#x27;s the first part of the ratio. And to our two times the radius is the diameter. So this can actually be rewritten as C over to or see over d d for diameter, since it&#x27;s the ratio of the circumference to the diameter. But for the purpose of the problem, this is the final solved answer.

Ifeyinwa Onyekwena · Answer

So we&#x27;re going to be comparing two formulas first of this circumference of a circle which has two times pi times the radius and then the area of a circle which has pie times the radius squared. So let&#x27;s think about what these two foremost having Harmon. Well, I see that they both have pie. So what I want to do now is factor out a pie from both of these formulates If I factor out pie from the form of this a conference I&#x27;m left with two times the radiance while if I factor pie from the former for area, I&#x27;m left with radius squared. So those are things they have in common. We have pie in common and the things that differentiate them as that&#x27;s the conference were more supplying two times the radius and with the area were squaring the radio one of filets that you can remember this as linger, trying to remember which formula has two multiplied in which formula has the radius squared. We can think about with the circumference and with the areas. So circumference is the distance around the circle. While the area is the space inside of a circle. Well, what happens with area and circumference is that area tends to be larger. And if you think about it, that makes sense. What&#x27;s inside the circle always will tend to be larger than the, um, circumference of the circle. Well, if you remember, the area should be larger. Then you can remember that. Okay, that&#x27;s the one where I square my radius. However, with us a conference since that tends to have a lower value, then you can remember. Okay, so that&#x27;s the one where he multiplied because most of wanting tips to give you a lower value been squaring.

Problem

MAKING AN ARGUMENT In the diagram, the measure of…

Problem 36 Hard Difficulty

HOW DO YOU SEE IT?
Compare the circumference of $\odot \mathrm{P}$ to the length of DE. Explain your reasoning.

Answer

The radius of $C$ is therefore 2$r .$ The arc length $D E$ is subtended by a
straight angle and so can be evaluated as: $\frac{180^{\circ}}{360^{\circ}} \times 2 \pi \times 2 r=2 \pi r .$ It can
therefore be seen that the arc length $D E$ is equal to the circumference of
the circle $P .$

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The radius of $C$ is therefore 2$r .$ The arc length $D E$ is subtended by astraight angle and so can be evaluated as: $\frac{180^{\circ}}{360^{\circ}} \times 2 \pi \times 2 r=2 \pi r .$ It cantherefore be seen that the arc length $D E$ is equal to the circumference ofthe circle $P .$

Geometry A Common Core Curriculum

Lily A.

Catherine R.

Maria G.

Michael D.

(A few) Circle Theorems

Angles and Cirlce Theorems with chords and tangents

Ron Larson, Laurie BoswellGeometry A Common Core Curriculum

Lily A.

Catherine R.

Maria G.

Michael D.

The radius of $C$ is therefore 2$r .$ The arc length $D E$ is subtended by a
straight angle and so can be evaluated as: $\frac{180^{\circ}}{360^{\circ}} \times 2 \pi \times 2 r=2 \pi r .$ It can
therefore be seen that the arc length $D E$ is equal to the circumference of
the circle $P .$

Ron Larson, Laurie Boswell

Geometry A Common Core Curriculum