Question

Find the area of the shaded region of the circle. Leave your answer in terms of π.

Answer 1

the first one i think is.
A = mA/360*pi r^2
A = 75 / 360 x pi x (9)^2
A= (135/8)*π

Answer 2

So, we'll use this formula to find the area of a sector.

\[\pi*(r^{2})*(\frac{ \theta }{ 360 }) = area~of~sector\]

Answer 3

pi*(2^2)*(120/360) = a

Answer 4

yeah, good work for the first one

\[\theta = 75^{o}~| \frac{ 75 }{ 360 }*81~(inches)^{2}~\pi = \frac{ 135 }{ 8 }\pi~inches^{2}\]

Answer 5

thanks

Answer 6

I have an idea. it's a little strange.

you see that triangle? the bas and the height are both the same

Answer 7

1/3*pi*2^2=A

Answer 8

yes i see it

Answer 9

I'm trying to use an alternative method to solve this problem.

You see what i'm thinking is that if we subtract the area of that triangle from the total area of the circle, we can find the area of the shaded region.



\[area~of~circle-(area~of~\triangle) = Area~shaded~region\]

Answer 10

okay

Answer 11

I'll try to do the problem two ways to see if we get the same answer using two different methods lol.


see I don't think arc length would help us out here and neither would the area of a whole sector. So, my thoughts are that the distance from the center of the circle is r so we've got a triangle of base and height r which we can find the area of.

we find the total area of the circle - the area of the triangle inscribed in the circle and we get that missing piece.

\[\pi*(r^{2})- \frac{ 1 }{ 2 }*(b*h) = Area~shaded~region\]

\[\pi*(2)^{2}-(\frac{ 1 }{ 2 })(2~m)^{2} = 4~\pi-2 = 3.36(\pi)\]

Answer 12

okay

Answer 13

Yeah we're done here.

Hey, @ganeshie8 I was just playing around with the formulas so see how theta and that 120 degrees are related.

say Circumference = 2pi*r

C = 2pi*r

and Arc length theta*r = L

Here's something else I was thinking. we know that the full circumference is 2*pi*r = C. so here, we've got 120 degrees. so i'm thinking that out of the arc length would be 1/3 of the full circumference 120/360*(4pi) or 1/3 of the circumference, which translates to 4/3pi



\[\frac{ 120 }{ 360 }*(4*\pi) = \frac{ 4 }{ 3 }\pi~ = L \]

now we can find theta.

\[\theta*r = L => \frac{ L}{ r } = \theta => \frac{ \frac{ 4 }{ 3 }\pi }{ 2 } = \frac{ 2 }{ 3 }\pi*(\frac{ 180 }{ \pi }) = 120^{o}\]

i'm wondering if we can form a relationship here.

my argument is that the 120 degrees written on the arc length. 120 degrees is actually the value of theta.