Let's do "b" first.....since this is an octagon. the measure of <AOB = 360/8 = 45
And m<AOX is 1/2 of that = 22.5
And that's a and b
c) Note that the measure of <OAX = 90- 22.5 = 67.5
And we can find OX by the Law of Sines
OX/sin67.5 = OA/sin90 ...... (sin 90 = 1)
OX/sin67.5 = 9
OX = 9sin67.5 = about 8.3m
d) We can find AB with the Law of Cosines
AB^2 = OA^2 + OB^2 - 2(OA)(OB)cos45
AB^2 = 9^2 + 9^2 - 2(9)(9)cos45
AB^2 = 162 - 162cos45 = 47.448701447706 take the square root of both sides
AB = about 6.9m
e) The perimeter = 8(6.9)= about 55.2 m
f) Area...we have 8 congruent triangles....the area of each is given by (1/2)(9^2)sin45 = 28.6378246380735m^2
So.... 8*28.6378246380735 = about 229m^2
Hope I haven't made any serious errors here.....!!!
Can i ask u CPhill whos Natasha!becoz ive just heard one name like this"Natasza"! so whos "Natasha "now!
I am Polish, So It Is Spellt Natasza! :) But American Spell it Natasha. Poor CPhill had Typoes with it! lol My Email Was Spellt Natasha so...Thats Why He Spell It Like That
oh thats why hes spelling it like that!but btw i have no problem spelling it i can simply say "Natasza" thats easy!and natasza r these questions by u!? i never saw u posting them!
Haha thats Okay! :) It is Pronounced Sort of The Same! These Questions Are Not Mine.
Let's do "b" first.....since this is an octagon. the measure of <AOB = 360/8 = 45
And m<AOX is 1/2 of that = 22.5
And that's a and b
c) Note that the measure of <OAX = 90- 22.5 = 67.5
And we can find OX by the Law of Sines
OX/sin67.5 = OA/sin90 ...... (sin 90 = 1)
OX/sin67.5 = 9
OX = 9sin67.5 = about 8.3m
d) We can find AB with the Law of Cosines
AB^2 = OA^2 + OB^2 - 2(OA)(OB)cos45
AB^2 = 9^2 + 9^2 - 2(9)(9)cos45
AB^2 = 162 - 162cos45 = 47.448701447706 take the square root of both sides
AB = about 6.9m
e) The perimeter = 8(6.9)= about 55.2 m
f) Area...we have 8 congruent triangles....the area of each is given by (1/2)(9^2)sin45 = 28.6378246380735m^2
So.... 8*28.6378246380735 = about 229m^2
Hope I haven't made any serious errors here.....!!!