1.) The area of a regular octagon is 35 cm squared. What is the area of a regular octagon with sides three times as long? My answer: 315 cm squared. Is that correct? 2.) Name the minor arc and find its measure. http://assets.openstudy.com/updates/attachments/4fb5f543e4b05565342bcb5a-kamuela710-1337324879047-4568453162011122648pm17916 My answer: ADB has a measure of 245. Is that correct?
3.) The figures below are similar. What are a) the ratio of the perimeters and b) the ratio of the areas of the larger figure to the smaller figure? The figures are not drawn to scale. http://assets.openstudy.com/updates/attachments/55bcef17e4b0c4d93d6c2874-lollygirl217-1438476949164-as.jpg My answer: 13/3=169/9. Is that correct?
@Michele_Laino @triciaal
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@amistre64
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@undeadknight26
how do we determine the area of a regular octagon to start with?
im pretty sure the general solution is that a change in linear measures, scale by ^2 in area
so 315 sounds good to me, but i always have to dbl chk with the brute method since i cant recall thrms that well
Alright, do your thing.
a triangle ... base s, apothem a, area = sa/2 and there is 8 of them area = 4 sa linear measures factor by 3 so: 4(3s)(3a) = 3^2 (4sa) in this case 3^2 (35) is good
the link to the arc is bad
Oh it is? Hold on.
ok, and what does it mean to be a minor arc?
@amistre64 well firsts things first. Did I get any of them right?
I mean that one right.
yes, i verified that 315 is correct. on the second one, and the third one, i would perfer to see how you came to your solutions first tho.
to determine the solution to number 2, we need to agree on the definition of a minor arc.
Perhaps it's just AB has a measure of 245.
AB is minor because of the two arcs in the circle AB has a measure of 115 degrees and ADB has a measure of 245. This is what I say
perhaps, but that depends on what our definition of a minor arc is ...
AB is minor, agreed, since it is less then 180 .. and it is measured for us already as 115
we good to move on to #3?
So then it would be ab measured as 115?
correct
Lol...how did I make that mistake?
:) pretty sure you read minor as "major" ... i read things too fast sometimes as well
Yeah, I did.
tell me how you approached #3
a) ratio of perimeters = 26/6 = 13/3 b) ratio of areas = (13/3)² = 169/9 <----------A = L²
i like it.
not sure if the notation is correct, since im not taking your class specifically 13:3, or 1 : 13/3 is what i am used to playing with
if the ratio of sides of similar shapes is: 1:k the ratio of their ares is: 1:k^2
So it is 13/3, I got have of it right. Then it must be 13/3 and 16/9
Or are you saying I got it right??
you got it right; 13/3 and 13^2/3^2 is fine
13^2 = 169 3^2 = 9 just make sure your notation is accurate to your course.
going from large to small might mean: 3/13 and 9/169
Alright thank you.
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