Question

Can someone please explain to me how this is wrong????

Answer 1

\(\bf \textit{area of a segment of a circle}=\cfrac{{\color{brown}{ r}}^2}{2}\left(\cfrac{{\color{brown}{ \theta}}r}{180}-sin({\color{brown}{ \theta}})\right)\)

Answer 2

I found the area of the hole circle which is 153.93804 then i multiplied that by 35 which is 5387.831 then i did 5387.831/360 which equals 14.96619?

Answer 3

Then i rounded to the nearest hole number which is 15

Answer 4

you're correct, is 14.9 which rounds to 15, yes, for the "sector" of the circle

Answer 5

Ok when i plug the numbers into the equation up there i get .787535? is that right?

Answer 6

https://www.google.com/search?client=opera&q=(7%5E2/2)*(((35*pi)/180)-sin(35+degrees))&sourceid=opera&ie=utf-8&oe=utf-8&channel=suggest&gws_rd=ssl

Answer 7

hmmm wait a sec... I missed something there

Answer 8

hmmm no .... shoot.. just have a typol lemme fix that quick

Answer 9

\(\bf \textit{area of a segment of a circle}=\cfrac{{\color{brown}{ r}}^2}{2}\left(\cfrac{{\color{brown}{ \theta}}\pi}{180}-sin({\color{brown}{ \theta}})\right)
\\ \quad \\
\implies \cfrac{{\color{brown}{ 7}}^2}{2}\left(\cfrac{{\color{brown}{ 35^o}}\cdot \pi}{180}-sin({\color{brown}{ 35^o}})\right)\)

Answer 10

as jdoe shows, the *segment* is the part of the sector cut off by a chord

Answer 11

thus https://www.google.com/search?client=opera&q=(7%5E2/2)*(((35*pi)/180)-sin(35+degrees))&sourceid=opera&ie=utf-8&oe=utf-8&channel=suggest&gws_rd=ssl

Answer 12

Another "formula" is to subtract off the area of the isosceles triangle with sides 7
and vertex angle 35º

(a bit complicated, because you must use trig, but it's easier to remember, than the formula posted up above)

Answer 13

Ok thanks

Answer 14

yw

Answer 15

the triangle has an altitude of 7 cos(35/2)
and base 2*7*sin(35/2)

and area 7*7*sin(35/2)*cos(35/2)

and if you know sin(2x) = 2 sin(x)cos(x)

½*7*7*sin(35)