The area of the regular octagon is calculated as half of the product of the perimeter and the apothem (ap), using the formula of the area of the regular polygon.
We have then:
A = ((p) * (ap)) / 2
Where,
p: perimeter
ap: apotema
Substituting values:
A = ((8 * 3.4) * (4.2)) / 2
A = 57.12 in ^ 2
Answer:
the area of the regular octagon is:
B. 57
2 Answers:
- Choice B) angle R and angle U
- Choice D) angle K and angle L
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Explanation:
There are a lot of lines here, so it's easy to get a bit lost if you're not careful. I recommend starting with a blank piece of paper. On it, draw two lines that intersect any way you want. An "X" shape forms. The angles that are opposite one another (eg: top and bottom angles) are known as vertical angles, and they are always congruent.
Going back to your diagram, angle R and angle U fit the description of vertical angles. We used two lines to form these vertical angles. Despite the name, the angles do not have to be aligned vertically. Unfortunately it's a misnomer in my opinion. Angle R is opposite angle U, and both are formed by the two slanted lines that go in both directions. It might help to copy the diagram your teacher has given to you, and then erase the portions that seem extra (eg: erase the ray that helps form angle S, and also erase angle S as well) and it might help show why angle R is a vertical angle pair with angle U.
Similarly, angle K and angle L fit the description as well, so this is the second pair of vertical angles.
Something like "angle E and angle N" is not an answer because we used 3 lines to help form these angles, instead of two only. We can rule out "angle S and angle J" for similar reasons.
As for when you see vertical angles in everyday life, it depends on what the context is. For example, if you're drawing out a floorplan, then knowing the angles may be very handy. You could use a protractor to measure the angles, or you could use geometric properties such as the vertical angle theorem to quickly determine the angle values (if you know some angles already).
Well, we know the cyclist left the western part going eastwards, at the same time the car left the eastern part going westwards
the distance between them is 476 miles, and they met 8.5hrs later
let's say after 8.5hrs, the cyclist has travelled "d" miles, whilst the car has travelled the slack, or 476-d, in the same 8.5hrs
we know the rate of the car is faster... so if the cyclist rate is say "r", then the car's rate is r+33.2
thus
![\bf \begin{array}{lccclll} &distance&rate&time\\ &-----&-----&-----\\ \textit{eastbound cyclist}&d&r&8.5\\ \textit{westbound car}&476-d&r+33.2&8.5 \end{array}\\\\ -----------------------------\\\\ \begin{cases} \boxed{d}=8.5r\\\\ 476-d=(r+33.2)8.5\\ ----------\\ 476-\boxed{8.5r}=(r+33.2)8.5 \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Blccclll%7D%0A%26distance%26rate%26time%5C%5C%0A%26-----%26-----%26-----%5C%5C%0A%5Ctextit%7Beastbound%20cyclist%7D%26d%26r%268.5%5C%5C%0A%5Ctextit%7Bwestbound%20car%7D%26476-d%26r%2B33.2%268.5%0A%5Cend%7Barray%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%5Cbegin%7Bcases%7D%0A%5Cboxed%7Bd%7D%3D8.5r%5C%5C%5C%5C%0A476-d%3D%28r%2B33.2%298.5%5C%5C%0A----------%5C%5C%0A476-%5Cboxed%7B8.5r%7D%3D%28r%2B33.2%298.5%0A%5Cend%7Bcases%7D)
solve for "r", to see how fast the cyclist was going
what about the car? well, the car's rate is r + 33.2